A ball is thrown vertically upwards at speed of 20 m s−1. Find the time taken by the ball to reach back to the original position. [Take acceleration die to gravity, g=10 m s−2]
In this question we have two cases,
In the first case the ball goes from the original position to a maximum height where its velocity becomes zero.
And in the second case the ball comes back from the top to the original position.
Calculating the time taken by the ball to reach the maximum height.
initial velocity, u=20 m s−1
final velocity, v=0
acceleration, a=−g=−10 m s−2 [acceleration is taken as negative as it is acting in a direction opposite to the direction of motion of the ball]
⇒t=−20−g=−20−10=2 sec ....(t1)
Similarly, in the second case when the ball is coming back from the top to the original position it has to cover same distance as that of the first case.
So for this case we have,
initial velocity, u=0 m s−1
final velocity, v=20 m s−1
acceleration, a=g=10 m s−2 [Positive acceleration]
⇒t=20g=2010=2 sec ....(t2)
So, the total time taken by the ball to reach the original position =t1+t2=2+2=4 s