What is the final downward velocity of a quarter dropped from a tower after 10 seconds?

Answer 1

If you assume the tower is tall enough that the quarter doesn't hit the ground before 10 seconds, you can use one of the kinematic equations of motion to solve this problem. Those equations are:

Δx = x_f - x_i

v_avg = ½ (v_f + v_i) = Δx / Δt

v_f = v_i + at

x_f = x_i + v_it + ½ at²

v_f² = v_i² + 2a(x_f − x_i)

In this case, since we are looking for velocity and we have the initial velocity (0, since it was dropped) and we have the time (10 sec), and we have the acceleration (the acceleration due to gravity is -9.81 m/s²), we would use the third equation v_f = v_i + at

v_f = 0 + (-9.81)(10) = -98.1 m/s (the minus just means its traveling downwards)