Question

A rock is dropped off of a very tall building. How fast is it going after it has fallen 13.4 m? Number Units How far has it fallen after a time of 2.84 s? Number Units

Use kinematic equations.

Answer 1

First part is based on third kinematic equation. v^{​​​​​​2}​​​​​ = u^{​​​​​​2} + 2gh

And 2nd part is based upon second equation of motion h = ut + ½gt^{​​​​2}

* Initial speed of the rock u = 0 height dropped h = 13.4 m final speed v = p acc^n 'g' = 9.8 m/s^2 hence using III^rd equation of motion v^2 = u^2 + 2 gh v^2 = 0 + 2 times 9.8 times 13.4 = 262.64 hence v = squareroot 262.64 = 16.2 m/s Ans * times t = 2.84 sec Distance fallen h = > hence using II^rd equation of motion h = ut + 1/2 gt^2 h = 0 times 2.84 + 1/2 (9.8 m/) (2.84) = 1/2 times 9.8 times (2.84)^2 h = 39.5 m Ans