Question

Two balls are dropped to the ground from different heights. One is dropped 1.5 s after the other, but they

both strike the ground at the same time, 5.0 s after the first was dropped. (a) What is the difference in the
heights from which they were dropped? (b) From what height was the first ball dropped? Ans. (a) 63 m;
(b) 0.12 km

Answer 1

Height difference: approximately
`63\; {\rm m}`.

The first ball was dropped from a height of approximately
`123\; {\rm m}`.

(Assumptions: both balls were released from rest, air friction is negligible, and that
`g = 9.81\; {\rm m\cdot s^(-2)}`.)

Step-by-step explanation:

Under the assumptions, both ball would accelerate at a constant
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}`.

Let
`t` denote the time since the first ball was released.

Height of the first ball at time
`t` can be modelled with the SUVAT equation
`h(t) = (1/2)\, a\, t^(2) + u\, t + h_(0)`, where
`u` is the initial velocity. However, since
`u = 0\; {\rm m\cdot s^(-1)}` by assumption, this equation simplifies to
`h(t) = (1/2)\, a\, t^(2) + h_(0)`.

Since this ball reached the ground after
`t = 5.0\; {\rm s}`,
`h(5.0) = 0\; {\rm m}`. In other words:

`\begin{aligned}(1)/(2)\, (-9.81)\, (5.0)^(2) + h_(0) = 0\end{aligned}`.

Simplify and solve for the initial height of this ball,
`h_(0)`:

`\begin{aligned}h_(0) &= -(1)/(2)\, (-9.81)\, (5.0)^(2) \\ &\approx 123\; {\rm m}\end{aligned}`.

In other words, the first ball was dropped from a height of approximately
`123\; {\rm m}`.

Similarly, the height of the second ball may be modelled as
`h(t) = (1/2)\, a\, t^(2) + h_(0)`.

Since this ball reached the ground
`t = (5.0 - 1.5)\; {\rm s} = 3.5\; {\rm s}` after being released,
`h(3.5) = 0\; {\rm m}`. The initial height of this ball would be:

`\begin{aligned}h_(0) &= -(1)/(2)\, (-9.81)\, (3.5)^(2) \\ &\approx (-60)\; {\rm m}\end{aligned}`.

Subtract the initial height of the second ball from that of the first ball to find the difference in initial height:

`(123 - 60) \; {\rm m} \approx 63\; {\rm m}`.