Question

If you were to throw a ball vertically upward with an initial velocity of $50m{s}^{-1}$, approximately how long would it take for the ball to return to your hand? Assume air resistance is negligible. (Given $g=10m{s}^{-2}$)

• A. $2.5s$
• B. $5.0s$
• C. $7.5s$
• D. $10s$

Answer 1

The correct option is D $10s$

The only force acting on the ball is the force of gravity. The ball will ascend until gravity reduces its velocity to zero and then it will descend. Find the time it takes for the ball to reach its maximum height and then double the time to cover the round trip.
Using $v=u+at=u-gt$, we get:
$0m/s=50m/s-\left(10m/s^2\right)t$
Therefore,
$t=\left(50m/s\right)/\left(10m/s^2\right)=5s$
This is the time it takes the ball to reach its maximum height. The total round trip time is $2t=10s$.