Answer to Question #128089 in Mechanics | Relativity for john

Question

A 50 g rock is launched vertically in the air by a slingshot. The slingshot exerts an average force of 53 N on the rock over a distance of 0.750 m.  

1. Find the maximum height reached by the rock in [m].
2. Find the speed of the rock as it leaves the slingshot in [m/s]?

Answer 1

We can calculate the total energy given by the slingshot to the rock in form of $E = \bar{F}\cdot S = 53\,\mathrm{N}\cdot 0.750\,\mathrm{m} = 39.75\,\mathrm{J}.$

2. This energy transforms into the kinetic energy of the rock, so the initial velocity can be calculated as $v = \sqrt{\dfrac{2E}{m}} = 39.9\,\mathrm{m/s}.$

1. The rock stops to rise when the kinetic energy turns to 0. So the final potential energy is equal to the initial kinetic energy. Therefore,

$mgh = \dfrac{mv^2}{2} = E, \;\; h = \dfrac{E}{mg} \approx 81\,\mathrm{m}.$