# Numerical Problems on Projectile Motion – 01

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#### Example – 01:

• A body is projected with the velocity of 49 m/s at an angle of 30o with the horizontal. Find a) the maximum height reached by the body, b) the time of ascent, c) the time of flight, d) horizontal range and e) The direction of its velocity after 1 s.
• Given: velocity of projection = vo = 49 m/s, Angle of projection = θ = 30o, g = 9.8 m/s2.
• To Find: Maximum height reached = H =?, Time of ascent = ta =?, time of flight = T =?, horizontal range = R =?, the direction of velocity = α =? At t = 1 s.
• Solution:

The maximum height reached is given by The time of ascent is given by The time of flight is given by The horizontal range is given by The direction of velocity after time t is given by Ans: Maximum height reached = 30.6 m, time of ascent = 2.5 s, time of flight = 5s

Horizontal range = 212.2 m, Direction of velocity after 1 s is 19o6’ with horizontal.

#### Example – 02:

• A body is projected with the velocity of 60 m/s at an angle of 30o with the horizontal. Find a) the maximum height reached by the body, b) the time of flight, c) horizontal range
• Given: velocity of projection = vo = 60 m/s, Angle of projection = θ = 30o, g = 9.8 m/s2.
• To Find: Maximum height reached = H =?, time of flight = T =?, horizontal range = R =?,
• Solution:

The maximum height reached is given by The time of flight is given by The horizontal range is given by Ans: Maximum height reached = 45.92 m, time of flight = 6.12 s, horizontal range = 318.1 m

#### Example – 03:

• A cricket ball is thrown at a speed of 28 m/s in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level
• Given: velocity of projection = vo = 28 m/s, Angle of projection = θ = 30o, g = 9.8 m/s2.
• To Find: Maximum height reached = H =?, time of flight = T =?, horizontal range = R =?,
• Solution:

The maximum height reached is given by The time of flight is given by The horizontal range is given by Ans: Maximum height reached = 10.0 m, time of flight = 2.86 s, horizontal range = 69.28 m

#### Example – 04:

• A body is projected with the velocity of 98 m/s at an angle of 60o with the horizontal. Find a) the maximum height reached by the body, b) horizontal range
• Given: velocity of projection = vo = 98 m/s, Angle of projection = θ = 60o, g = 9.8 m/s2.
• To Find: Maximum height reached = H =?, horizontal range = R =?,
• Solution:

The maximum height reached is given by The horizontal range is given by Ans: Maximum height reached = 367.5 m, horizontal range = 848.7 m

#### Example – 05:

• A shell is projected with the velocity of 196 m/s at an angle of 30o with the horizontal. Find a) the maximum height reached by the shell, b) the time of flight, c) horizontal range
• Given: velocity of projection = vo = 196 m/s, Angle of projection = θ = 30o, g = 9.8 m/s2.
• To Find: Maximum height reached = H =?, time of flight = T =?, horizontal range = R =?,
• Solution:

The maximum height reached is given by The time of flight is given by The horizontal range is given by Ans: Maximum height reached = 490 m, time of flight = 20 s, horizontal range = 3395 m

#### Example – 06:

• A body is projected with the velocity of 40 m/s at an angle of 30o with the horizontal. Find a) the maximum height reached by the shell, b) the time taken to reach maximum height, c) horizontal range
• Given: velocity of projection = vo = 40 m/s, Angle of projection = θ = 30o, g = 9.8 m/s2.
• To Find: Maximum height reached = H =?, time of ascent  = ta =?, horizontal range = R =?,
• Solution:

The maximum height reached is given by The time to reach maximum height is given by The horizontal range is given by Ans: Maximum height reached = 20.41 m, time required to reach maximum height = 2.041 s, horizontal range = 141.4 m

#### Example – 07:

• A bullet is fired from a rifle attains a maximum height of 25 m and a horizontal range of 200 m. Find the angle of projection.
• Given: maximum height reached = H = 25 m, horizontal range = R = 200m, g = 9.8 m/s2.
• To Find: Angle of projection = θ =?
• Solution:

The maximum height reached is given by The horizontal range is given by Dividing equation (1) by (2) Ans: The angle of projection is 26o34’

#### Example – 08:

• A projectile attains a maximum height of 100 m and a horizontal range of 400 m. Find the angle of projection and velocity of projection.
• Given: maximum height reached = H = 100 m, horizontal range = R = 400m, g = 9.8 m/s2.
• To Find: Angle of projection = θ =?, velocity of projection = vo =?
• Solution:

The maximum height reached is given by The horizontal range is given by Dividing equation (1) by (2) The range of a projectile is given by Ans: The angle of projection is 45o and the velocity of projection is 62.61 m/s

#### Example – 09:

• The horizontal range of a projectile is equal to the maximum height attained by the body. What is the angle of projection?
• Given: maximum height reached (H) = horizontal range (R), g = 9.8 m/s2.
• To Find: Angle of projection = θ =?
• Solution:

The maximum height reached is given by The horizontal range is given by Given R = H Ans: The angle of projection is 75o48’

#### Example – 10:

• The maximum horizontal range of a projectile is 980 m. Find the velocity of its projection.
• Given: maximum horizontal range = R = 980 m, g = 9.8 m/s2.
• To Find: velocity of projection  = vo =?
• Solution:

For maximum range θ = 45o

The horizontal range is given by Ans: Velocity of projection is 98 m/s

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