Question

how long will it take for a ball to hit the ground if it's thrown directly up at 45 m/s on a level surface? Off a 150 m cliff?

Answer 1

To reach the ground:

initial velocity u = 45 m/s

Acceleration due to gravity = -9.8 m/s^2

We know the kinematic equation

x = distance travelled, a = acceleration, t = time

Here x = 0 m (ball reaches to the initial position), a = -9.8 m/s^2

substituting in equation 1 we get time

t = 9.18 seconds

case 2 from a 150 m cliff:

From the cliff 150 m above ball is thrown upwards, similarly the time taken we calculated as above is 9.18 seconds.

The height covered by the ball can be calculated using

v = final velocity , that is zero at maximum height x,

solving we get x = 103 metres from the cliff.

Therefore the total distance from the ground is 150 metres + 103 metres = 253 metres.

Now we need to calculate the time required to reach ground from the cliff using equation 1

on solving for t we get

T = 7.18 seconds.

Therefore the total time in travel is 7.18 sec + 9.18 sec = 16.36 seconds.