Question

Suppose that a ball is dropped from the upper observation deck of the CN tower, 450 m above the ground.

b)How fast is the ball traveling when it hits the ground?
Thanks.

Answer 1

We'll want to use the following equation from kinematics:
\[v_f ^2 = v_i ^2 + 2ad\] Where v_f and v_i are the final and initial velocities, a is acceleration and d is distance. The ball starts from rest so v_i=0. So:
\[v_f ^2 = 2ad\]Plug in our acceleration due to gravity and distance:
\[v_f ^2 = 2(9.8 m/s^2)(450 m)\]And solve for v_f.