Question

A ball is thrown upward. After reaching a
maximum height, it continues falling back toward
Earth. On the way down, the ball is
caught at the same height at which it was
thrown upward.
v0
hmax
If the time (up and down) the ball remains
in the air is 1.84 s, find its speed
when it caught. The acceleration of gravity
is 9.8 m/s
2
. Neglect air resistance.
Answer in units of m/s.

Answer 1

Answer:

9.0 m/s

Explanation:

The problem can be solved by using the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

t is the time

When the ball reaches the maximum height, the velocity becomes zero. If we consider this as initial point of the motion, we can write

u = 0

The time it takes for the ball to go down from the maximum heigth to the ground is half the total time, so

$t=\frac{1.84s}{2}=0.92 s$

And solving the formula for v, we find the final velocity:

$v=0+(-9.8)(0.92)=-9.0 m/s$

So, the final speed is 9.0 m/s.