Velocity physics help

Question

A boy throws a ball up in the air with a velocity of 7 m/s when they are standing at the edge of a cliff 20 m high. If the ball falls and lands at the base of the cliff, what is the velocity of the ball when it lands? How long was the ball in the air?

Answer 1

This question requires knowledge of the kinematic equations in order to solve. Here are the ones I used to solve this problem:

(1) x_f = x_i + v_it + 0.5at²

x_f = Final position after time t

x_i = Initial position at time t = 0

v_i = Initial velocity at time t = 0

a = Acceleration on the ball during flight (constant)

t = Time in which the ball traveled from x_i to x_f

(2) v_f = v_i + at

v_f = Final velocity after time t

v_i = Initial velocity at time t = 0

a = Acceleration on the ball during flight (constant)

t = Time in which the ball traveled from x_i to x_f

Step 1: Define what you know

x_f = 0m (Bottom of cliff)

x_i = 20m (Top of cliff where ball was thrown)

v_i = 7m/s

a = -9.81 m/s² (Acceleration due to gravity; value is negative since we are defining up as the positive direction and down as the negative direction)

Step 2: Plug into Equation 1 and solve for t using quadratic formula. Throw away answer that has a negative time (since that can't exist) and the positive value is your answer for t.

0 = 20 + 7t - 4.9t²

t = -1.43; t = 2.86s

Step 2: Plug into Equation 2 and solve for v_f.

v_f = 7 - 9.81(2.86)

v_f = -21.06m/s (the negative is from us defining down as the negative direction. This makes sense intuitively since at the bottom of the cliff, the ball is traveling towards the ground)

v_f = 21.06m/s

Answer 2