Question

What is initial velocity of a ball (m/s) which is thrown upwards and takes 15.0s to return to its starting position?

A. 73.6 m/s
B. 73.6 m/s^2
C. 25 m/s^2
D. 25 m/s

Show the work for determining the velocity of the ball...show symbolic solution then numerical solution.

Answer 1

Approximately
`73.6\; {\rm m\cdot s^(-1)}`.

(
`v = (1/2)\, g\, t`, assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`.)

Step-by-step explanation:

Assume that the air resistance on the ball is negligible. Let
`v` denote the initial velocity of the ball.

The kinetic energy of the ball will be conserved. Hence, when the ball returns to the starting position, the ball will be travelling at the same speed but in the opposite direction (downwards.) The velocity will become
`(-v)`.

The change in the velocity of the ball would be
`\Delta v = ((-v) - v) = (-2\, v)`.

Change in velocity is also equal to
`a\, t`, where
`a` is acceleration and
`t` is the time required to achieve such change. Under the assumptions, acceleration of the ball will be constantly
`a = (-g)`. Hence:

`\Delta v = a\, t = (-g)\, t`.

Since
`\Delta v = ((-v) - v) = (-2\, v)`:

`(-2\, v) = \Delta v = (-g)\, t`.

`\begin{aligned}v &= ((-g)\, t)/((-2)) = (g\, t)/(2)\end{aligned}`.

Substitute in
`g = 9.81\; {\rm m\cdot s^(-2)}` and
`t = 15.0\; {\rm s}` to obtain:

`\begin{aligned}v &= (g\, t)/(2) \\ &= ((9.81)\, (15.0))/(2)\; {\rm m\cdot s^(-1)} \\ &\approx 73.6\; {\rm m\cdot s^(-1)}\end{aligned}`.