Question

7. How much time would it take a bowling ball dropped from a helicopter to reach a speed of 55 m/sec?

How far will the ball have fallen in that time?

Answer 1

Approximately
`5.61\; {\rm s}`.

Approximately
`154\; {\rm m}`.

(Assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`, air resistance on the ball is negligible, and that the helicopter was initially stationary.)

Step-by-step explanation:

Under the assumptions, acceleration of the ball during the fall would be constantly
`a = g = 9.81\; {\rm m\cdot s^(-2)}`.

• Let
  `u` denote the initial velocity of the ball;
  `u = 0\; {\rm m\cdot s^(-1)}` since it is assumed that the helicopter was initially stationary.
• Let
  `v` denote the final velocity of the ball;
  `v = 55\; {\rm m\cdot s^(-1)}`.

Apply the SUVAT equation
`t = (v - u) / a` to find the time it takes for the velocity of the ball to change from
`u = 0\; {\rm m\cdot s^(-1)}` to
`v = 55\; {\rm m\cdot s^(-1)}`:

`\begin{aligned}t &= (v - u)/(a) \\ &= (55 - 0)/(9.81)\; {\rm s} \\ &\approx 5.61\; {\rm s}\end{aligned}`.

Apply the SUVAT equation
`x = (v^(2) - u^(2)) / (2\, a)` to find the displacement
`x` of the ball as its velocity change from
`u = 0\; {\rm m\cdot s^(-1)}` to
`v = 55\; {\rm m\cdot s^(-1)}`:

`\begin{aligned}x &= (v^(2) - u^(2))/(2\, a) \\ &= (55^(2) - 0^(2))/(2\, (9.81))\; {\rm m} \\ &\approx 154\; {\rm m}\end{aligned}`.