Question

A rock is launched vertically upward by a slingshot. The elastic band of the slingshot accelerates the rock from rest to 40 m/s in a distance of 10 cm. How long will the rock take to return to its release point (time of flight)? Neglect air resistance.

Answer 1

Answer:

0.016 s

Explanation:

Initial velocity, u = 0

use the third equation of motion to find the acceleration of the rock.

$2as=v^2-u^2\\a = \frac{40^2-0}{2\times0.1} =8000 m/s^2$

Net displacement in the vertical direction would be zero. y = 0

Use second equation:

$s=ut+\frac{1}{2}at^2\\0=0+\frac{1}{2}(8000)t^2\\t= 0.016 s$

Thus, the rock will return to its initial release point in 0.016 s.

Answer 2

The rock takes 8.16s to return to its release point.

Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

Initial speed of the rock, u = 40m/s

Final position of the rock s = 0m taking the release point as reference

From the second equation of motion:

$s =ut-\frac{1}{2}gt^2 \\\\0=40t-\frac{1}{2}9.8t^2\\\\ t(40-4.9t)=0$

solving above we get:

t = 0s or t = 8.16s,  t =0  seconds is neglected since it represents the initial position which is the same as the final position at t = 8.16s

So, the rock takes 8.16 seconds to return to the release point.

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