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

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.