**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-bx^{2 }as 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 along_{x }along the x-direction and u_{y }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

**Where**

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

**Where**

At the same time the horizontal and vertical displacements

can be obtained from the second equation linear motion

For horizontal displacement

Since

For vertical displacement

Given that

(Shows that the acceleration is against the gravity)

**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 =

. Substitute for t in (3.6)

**On simplification we have**

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 v_{y} = 0 as shown in the figure 3.3.

**Figure3.3: Maximum height and range of a projectile**

From equation (3.4) when

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

When simplified this expression become

**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.

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.

Either

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)

**When**

R = u^{2} sin^{2}θ/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

**Rearrange the expression**

R = U^{2} (2cosθsinθ)/g

From the trigonometric identities

∴ R = U^{2} sin^{2}θ)/g

For maximum range

**Example**

**1:** A projectile is launched from a point on a level ground with a velocity of in 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.** **

**Figure 3.2**

**(b) The initial velocities **

(i)

(ii)

(**c) The velocities **

**(d) The greatest height Y**

(**e) The time of flight**

**(f) The range R**

**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.

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 *

**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

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

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

(a) To find

and

Let t = time taken

From

(b) By the moment it lands on the ground, the vertical distance covered is

**From**

-It takes the bullet 4.948 seconds to land on its target

(c) Since the horizontal velocity is constant through out the motion

**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/s^{2}

**Solution**

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

From

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

(i)

(ii)

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

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

(c) Let

By forming the triangle of vector we can use Pythagoras theorem

The velocity of the shell is 153 m

The direction of the shell at this moment is given by

**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,

Where

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

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}.

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 to the horizontal as in figure 3. When it reaches its peak, it has,

coordinates given by

and when it strikes the ground, its coordinates are

, where R is the horizontal range,

(a) Show that it reaches the maximum height, Y, given by

(b) Show that its horizontal range is given by

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

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.

**Figure 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.