Kinematics- Physics

Question

A ball is thrown upwards at 25 m/s at the edge of a cliff. It hits the ground at the

bottom of the cliff 12 seconds after being released. SHOW ALL WORK (equation,

algebraic rearrangement, substitution with units, and answer with units).

(a) Find the ball’s maximum height above the edge of the cliff.

(b) How high is the cliff?

(c) After 7.0 seconds, what is the velocity of the ball?

Answer 1

Step 1: Setup the kinematic equation with respect to the y-direction.

x_f = x₀ + v₀t + 0.5at²

x_f = Final position of ball = 0m

x₀ = Initial position of ball

v₀ = Initial velocity of ball = 25m/s

t = time = 12sec

a = Acceleration = -9.81m/s²

Solve for x₀

x₀ = 405.6m (how high is cliff)

Let's also find the height of the ball at its highest point. This will be when the instantaneous velocity in the y-direction is zero.

v_f = v₀ + at

v_f = 0m/s

v₀ = 25m/s

a = -9.81m/s²

Solve for t

t @ highest point = 2.55s

Use this t to find the height of the ball at its highest point.

x_f = x₀ + v₀t + 0.5at²

x_f = Final position of ball

x₀ = Initial position of ball = 405.6m

v₀ = Initial velocity of ball = 25m/s

t = time = 2.55sec

Solve for x_f

x_f = 437.5m

Find the ball’s maximum height above the edge of the cliff by subtracting x_f from x₀

437.5 - 405.6 = 31.9m

Use same equation to find the ball's velocity after 7 seconds.

v_f = v₀ + at

v₀ = 25m/s

a = -9.81m/s²

t = 7s

Solve for v_f

v_f = -43.7m/s