Question

A body is thrown vertically upwards with an initial velocity of 21m/s. How high will it go and when will it return to the point of projection? Assume the value of acceleration due to gravity is 10m/S.

Answer 1

To find the height reached by the body and the time it takes to return to the point of projection, we can use the equations of motion for vertical motion under constant acceleration.

Given: Initial velocity, u = 21 m/s (upwards) Acceleration due to gravity, g = 10 m/s² (downwards)

1. To find the height reached (maximum height), we can use the equation: v² = u² + 2as

At the highest point, the final velocity (v) will be 0 m/s. So, substituting the values into the equation: 0 = (21)² + 2(-10)s

441 - 20s = 0 20s = 441 s = 22.05 m

Therefore, the body will reach a maximum height of 22.05 meters.

2. To find the time taken to return to the point of projection, we can use the equation: v = u + at

At the highest point, the velocity is 0 m/s. So, substituting the values into the equation: 0 = 21 + (-10)t

-10t = -21 t = 2.1 seconds

Therefore, the body will take 2.1 seconds to return to the point of projection.