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PROJECTILE MOTION PHYSICS FORM 5

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PHYSICS ADVANCED LEVEL FULL NOTES FORM 5 AND 6 NEWTON’S LAWS OF MOTION-COLLISION NEWTON’S LAWS OF MOTION - MOTION PROJECTILE MOTION PHYSICS FORM 5 CIRCULAR MOTION PHYSICS FORM 5 ERRORS PHYSICS FORM 5 MEASUREMENT PHYSICS FORM 5

PROJECTILE MOTION PHYSICS FORM 5

A projectile is any object that when given an initial velocity it moves freely in space under the influence of gravity. The path it follows is known as trajectory which is in the form of a parabola y = ax-bxas shown in fig 3.1

Projectile motion is common in warfare, sports, hunting fire fighting and irrigation. To describe the motion of a projectile we shall make use of the equation of linear motion.

This is because a projectile moves in two directions simultaneously, that is, it moves along the x-direction and along the y-direction.

Initial velocities at the point of projection

Suppose a projectile is projected with an initial velocity u along the direction making an angle to the horizontal plane as in fig 3.1.

This velocity is divided into two components alongx along the x-direction and uy along the y-direction. Fig 3.1 shows how these components can be obtained by first forming the triangle of velocities and then using the trigonometric ratios of the right-angled triangle.

From the right angled -triangle in fig 3.2 we can establish the expressions of the initial velocities as follows:
For x – direction

For y-direction

After a time t, the velocities of the projectile  can be obtained from the first equation of linear motion

Along the x-direction

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image008.Gif

Where

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image009.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image010.Gif

This means that at any time t when the projectile is in flight the horizontal component of the initial velocity remains constant.

Along the y-direction

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image011.Gif

Where

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image012.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image013.Gif

At the same time the horizontal and vertical displacements C:\Thlb\Cr\Tz\Projectilemotion_Files\Image014.Gif

can be obtained from the second equation linear motion  C:\Thlb\Cr\Tz\Projectilemotion_Files\Image015.Gif

For horizontal displacement

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image016.Gif

Since

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image017.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image018.Gif

For vertical displacement

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image019.Gif

Given that C:\Thlb\Cr\Tz\Projectilemotion_Files\Image012.Gif

(Shows that the acceleration is against the gravity)

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image020.Gif

Equation of a trajectory

The  mathematical relationship between horizontal and vertical displacements that defines the path followed by the projectile.
This can be obtained by combining-equations (3.5) and (3.6) as follows:

From equation, (3.5) t = C:\Thlb\Cr\Tz\Projectilemotion_Files\Image021.Gif

. Substitute for t in (3.6)

C:\Thlb\Cr\Tz\__I__Images__I__\Let.jpg

On simplification we have

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image023.Gif

From the above equations it is now possible to get other expressions for the maximum height, time of flight, range and maximum range.

Maximum height

The maximum height is the vertical distance above the horizontal plane a projectile can possible attain for given initial velocity and angle of projection.

This is the position where a projectile ceases to move upwards in This case vy = 0 as shown in the figure 3.3.
C:\Thlb\Cr\Tz\__I__Images__I__\Eddy1021.Jpg

Figure3.3: Maximum height and range of a projectile

From equation (3.4) when C:\Thlb\Cr\Tz\Projectilemotion_Files\Image025.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image026.Gif

This is the time taken by the projectile to attain the maximum height γ. Substituting for t in equation (3.6) we get

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image027.Gif

When simplified this expression become

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image028.Gif

Time of flight

The time taken by the projectile to move from the point of projection to the target’ horizontally is known as time of flight. In other words it is the time a projectile remains in air.

See also  SIMPLE PRESENT TENSE

When the projectile lands on the target, the vertical displacement becomes zero which means y – 0. Using this condition in equation (3.6) we can solve for t as follows.

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image029.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image030.Gif

Either  C:\Thlb\Cr\Tz\Projectilemotion_Files\Image031.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image032.Gif

Thus the time of flight is twice the time taken by projectile to reach the maximum height.

The Range (R)

The range of a projectile is the horizontal displacement between the point of projection and the target. Since the horizontal component of the initial velocity of the projectile remains constant, the range is obtained from equation (3.5)

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image033.Gif

When

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image034.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image035.Gif

C:\Thlb\Cr\Tz\__I__Images__I__\Nit39.Png

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image036.Gif

R = u2 sin2θ/g … … … … … … … (3.9).

The Maximum Range

When a projectile is projected at different angles with the horizontal, the ranges covered differ. However there is a single angle for which the range is greatest of all. We can get this angle from the range expression in equation (3.9).

Write the range expression as

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image037.GifRearrange the expression

R = U2 (2cosθsinθ)/g

From the trigonometric identities

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image039.Gif

∴  R = U2 sin2θ)/g

For maximum range C:\Thlb\Cr\Tz\Projectilemotion_Files\Image041.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image042.Gif

Example

1: A projectile is launched from a point on a level ground with a velocity of C:\Thlb\Cr\Tz\Projectilemotion_Files\Image043.Gifin the direction making an angle of 60° to the horizontal.

(a)  Sketch the trajectory of the projectile

(b)  What are the initial velocities of the projectile along the x- and y-directions?

(c) Find the velocities of the projectile along the horizontal and vertical planes 3seconds after launch

(d)  Determine the greatest height attained by the projectile

(e) What is the time of flight?

(f)  What is the range?

Solution

The main assumptions to consider in a projectile motion are such as the effect of air resistance is reflected the effect of earth curvature  and it is rotation are neglected. 

C:\Thlb\Cr\Tz\__I__Images__I__\4412.PngFigure 3.2

C:\Thlb\Cr\Tz\__I__Images__I__\Let11Phyf5.Jpg

(b) The initial velocities  

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image045.Gif(i)    C:\Thlb\Cr\Tz\Projectilemotion_Files\Image046.Gif

C:\Thlb\Cr\Tz\__I__Images__I__\Nit41.PngC:\Thlb\Cr\Tz\Projectilemotion_Files\Image048.Gif(ii)      C:\Thlb\Cr\Tz\Projectilemotion_Files\Image049.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image050.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image051.Gif

(c)  The velocities  

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image052.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image053.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image054.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image055.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image056.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image057.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image058.Gif

(d) The greatest height Y

C:\Thlb\Cr\Tz\__I__Images__I__\Nit42.Png

C:\Thlb\Cr\Tz\__I__Images__I__\Nit43.Png

C:\Thlb\Cr\Tz\__I__Images__I__\Nit44.Png

(e) The time of flight

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image061.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image062.Gif

(f) The range R

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image063.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image064.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image065.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image066.Gif

Projectile fired from raised ground

Let us consider the projectile fired from the top of a cliff. Here the projectile may be launched horizontally or at an angle with the horizontal.

The equations used to describe the motion in this case are the same as what have been derived so far but with some modifications.

C:\Thlb\Cr\Tz\__I__Images__I__\Let2Phyf5Project.jpg

Figure 3.4 projectile fired horizontally

Projectile fired horizontally from the top of a cliff 

Consider a projectile launched horizontally with velocity u from the top of a cliff as shown in fig 3.4. The gravity causes the projectile to follow the trajectory as it advances the nature of which is identical to that in fig 3.1. The following are the factors to observe:

The horizontal velocity remains constant throughout the motion

The initial velocity along the horizontal direction u = ux and that along the vertical direction Uy =  0

The vertical distance projectile fall through is always assigned negative sign.

The time the projectile  takes to move along the trajectory equal to the time it takes to fall through the vertical distances  h

The resultant velocity v is obtained by Pythagoras’s theorem  C:\Thlb\Cr\Tz\__I__Images__I__\4413.Png

See also  6: ELEMENTARY ASTRONOMY | PHYSICS FORM 4

Example 2

A bullet is fired horizontally from the gun with muzzle velocity of 200 ms1from the top of a cliff 120m high.

Find.
(a)  The velocities of the bullet after it has fallen  C:\Thlb\Cr\Tz\__I__Images__I__\Nit46.Png

of the vertical distance

(b)   The time it takes to land on a level ground

(c)   The distance from the foot of a cliff to where it lands

C:\Thlb\Cr\Tz\__I__Images__I__\Lpp1.Jpg

Figure. 3.4 A bullet  fired from top of the cliff.

(a) To find C:\Thlb\Cr\Tz\Projectilemotion_Files\Image069.Gif

and C:\Thlb\Cr\Tz\Projectilemotion_Files\Image070.Gif

Let t = time taken

From
C:\Thlb\Cr\Tz\Projectilemotion_Files\Image071.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image072.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image073.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image074.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image075.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image076.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image077.Gif

C:\Thlb\Cr\Tz\__I__Images__I__\Nit47.PngC:\Thlb\Cr\Tz\Projectilemotion_Files\Image079.Gif

(b) By the moment it lands on the ground, the vertical distance covered is C:\Thlb\Cr\Tz\Projectilemotion_Files\Image080.Gif

From
C:\Thlb\Cr\Tz\Projectilemotion_Files\Image071.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image081.GifC:\Thlb\Cr\Tz\Projectilemotion_Files\Image082.GifC:\Thlb\Cr\Tz\Projectilemotion_Files\Image083.GifC:\Thlb\Cr\Tz\__I__Images__I__\Nit48.Png

-It takes the bullet 4.948 seconds to land on its target
(c)  Since the horizontal velocity is constant through out the motion
C:\Thlb\Cr\Tz\Projectilemotion_Files\Image086.Gif

C:\Thlb\Cr\Tz\__I__Images__I__\Nit49.Png

Example 3

A shell is launched from the top of a hill 90m above the plane ground with a velocity of 150 ms-1 in the direction making an angle 30° to the horizontal. Find

(a) The velocities along the x- and y-directions when it is half-way between the top and ground

(b) The horizontal distance covered by the moment it lands on the ground

The velocity and direction as the shell hits the ground. (Use g = 9.8m/s2

Solution

(a) Let t = time for the shell to descend 45m

From

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image071.Gif

C:\Thlb\Cr\Tz\__I__Images__I__\Nit50.Png

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image089.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image090.Gif

This is a quadratic equation in t that can be solved by the formula

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image092.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image093.Gif

C:\Thlb\Cr\Tz\__I__Images__I__\Nit53.Png

C:\Thlb\Cr\Tz\__I__Images__I__\Nit54.Png

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image096.Gif

(i)  C:\Thlb\Cr\Tz\Projectilemotion_Files\Image097.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image098.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image099.Gif

(ii)  C:\Thlb\Cr\Tz\Projectilemotion_Files\Image100.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image101.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image102.Gif

(b) When it lands on the plane the distance covered is now -90 m and therefore from

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image071.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image103.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image104.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image105.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image106.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image107.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image108.Gif

The shell takes 16.4s to the land on the target and by then the horizontal distance is

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image086.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image109.Gif

(c) Let  C:\Thlb\Cr\Tz\Projectilemotion_Files\Image110.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image111.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image112.Gif

By forming the triangle of vector we can use Pythagoras theorem

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image113.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image114.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image115.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image116.Gif

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image117.Gif

The velocity of the shell is 153 m C:\Thlb\Cr\Tz\Projectilemotion_Files\Image118.Gif

The direction of the shell at this moment is given by

C:\Thlb\Cr\Tz\Projectilemotion_Files\Image119.Gif

Exercise

1. A projectile is launched from a point on a horizontal ground with an initial velocity of 100ms”1 along the direction making an angle of 60° with the horizontal.

a) Sketch a labeled diagram showing how the projectile moves.

(b) What are the initial velocities at the point of projection

(c)  Determine the velocities of the projectile 4 seconds after projection. what is the highest point does the projectile reach while in motion?

d) Calculate the range of the projectile.

2 (a) Show that the equation of the trajectory of a projectile is,
C:\Thlb\Cr\Tz\__I__Images__I__\Nit55.Png

Where C:\Thlb\Cr\Tz\Projectilemotion_Files\Image121.Gif

angle of projection, u= initial velocity, x =horizontal distance, y = vertical distance and g = acceleration due to gravity.

(b) State the condition for a projectile to attain the maximum range and show that the maximum range is C:\Thlb\Cr\Tz\Projectilemotion_Files\Image122.Gif

3. Find the two possible angles of projection for a projectile to just clear the wall 10m high if the point of projection is 40m from the wall and the initial velocity of a projectile is 50ms-1.

See also  BIOLOGY FORM 6 NOTES ALL TOPICS

4. A man standing on a cliff 50m high sees a dog running away 20m from the footnote cliff. He throws a stone horizontally with velocity of 30ms-1, if the stone hits the dog, find

(a) The distance of the dog from the cliff

(b) The speed of the dog by the time it is hit by the stone.(Take g = 9.8ms-2)

5. A projectile is fired with an initial velocity of u at an angle C:\Thlb\Cr\Tz\Projectilemotion_Files\Image123.Gifto the horizontal as in figure 3. When it reaches its peak, it has, C:\Thlb\Cr\Tz\Projectilemotion_Files\Image124.Gif

coordinates given by C:\Thlb\Cr\Tz\Projectilemotion_Files\Image125.Gif

and when it strikes the ground, its coordinates are C:\Thlb\Cr\Tz\Projectilemotion_Files\Image126.Gif

, where R is the horizontal range,
(a) Show that it reaches the maximum height, Y, given by  C:\Thlb\Cr\Tz\Projectilemotion_Files\Image127.Gif

(b) Show that its horizontal range is given by C:\Thlb\Cr\Tz\Projectilemotion_Files\Image128.Gif

6 In fig 3.6, a projectile is fired at a falling target. The target begins falling at the same time as projectile leaves the gun. Assuming that the gun is initially aimed at the target, show that the target will be hit.

7. An object slides from rest along a frictionless roof 8m long, inclined 37°to the horizontal. While sliding, the object accelerates at C:\Thlb\Cr\Tz\__I__Images__I__\4415.Png

towards the edge of the roof which is 6m above the ‘ground. Find (a) the velocity components when it reaches the edge of the roof (b) the total time it remains in motion (c) the distance from the wall of the house to where the object hits the ground.

C:\Thlb\Cr\Tz\__I__Images__I__\Nit57.PngFigure 3.6

8.  A particle is projected at point on a level ground in such a way that its horizontal and
vertical components of the initial velocity are 30m/s and 30m/s respectively. From this
information ;find.

(a)    The highest point above the ground reached, by the particle

(b)   The horizontal distance covered after landing on the ground

(c)    The magnitude and direction of the initial velocity.

9. A warplane flying horizontally at l000kmh-1 releases a bomb at a height of 1000m.| The bomb hits the intended target, what was the distance of the plane from the target when the bomb was released?

10.A basketball player 1.8m tall throws a ball at a velocity of 10m/s in the direction 40º with the horizontal. The ball passes through the basket fixed at a height of 3m above the ground level. Find the horizontal distance of the basket from the point where the ball was thrown.

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