Question

A ball is thrown vertically upward from the edge of a bridge 22.0 m high with an initial speed of 16.0 m/s. The ball falls all the way down and strikes the water below. Determine the magnitude of the velocity of the ball just before it strike the water.

Answer 1

To solve this problem we will apply the concepts related to the kinematic equations of motion. We will start calculating the maximum height with the given speed, and once the total height of fall is obtained, we will proceed to calculate with the same formula and the new height, the speed of fall.

The expression to find the change in velocity and the height is,

$v_f^2-v_0^2 = -2gh$

Replacing,

$0^2-16^2 = -2(9.8)h$

$h = 13.0612m$

Thus the total height reached by the ball is

H = 22m+13.0612m

H = 35.0612m

Now calculate the velocity while dropping down from the maximum height as follows

$v_f^2-v_0^2 = 2gh$

Substituting the new height,

$v_f^2 - 0^2 = 2(9.8)(35.0612)$

$v = \sqrt{687.2}$

$v = 26.2145m/s$