Dennis drops a 6.0g quarter from the top of a building . If the quarter hits the ground moving at 30m/s ,then how high up is the top of the building ? (Ignore air drag )

Answer 1

This gets tricky if you don't have the time given, but you do have the final velocity and you have the gravitational constant of g = 9.8 m/s² and from these you can get what you need.

Freefall equation:

H = 1/2 gt²

But you don't have t. Here's how to derive it: if v_final = gt, then t = v_final / g

So substitute for t and plug and chug.

H = 1/2 g (v_final/g)² = 1/2 (9.8) (30/9.8)² = 15 m tall building

Hope that helps.

Answer 2

You can solve this as a kinematics problem under constant acceleration of gravity.

y = distance fallen under acceleration of gravity

v₀ = initial speed = 0 (released from rest)

v₁ = final speed = 30 m/s

g = acceleration of gravity = 9.8 m/s²

v₁² - v₀² = 2gy

v₁² = 2gy

y = v₁² / (2g) = 30² / [2(9.8)] m = ? meters

Note the answer does not depend on the mass of the object.