H2 Physics 5 - Projectile Motion and Energy: Ball Launched from a Cliff

Step 1 - Set up the problem

A ball is launched horizontally from a 45 m cliff at 12 m/s.

Step 1 - Set up the problem

Neglect air resistance. Find the time of flight, horizontal range, and speed at impact.

Step 1 - Set up the problem

Take g = 9.81 m/s squared.

Step 2 - Find the time of flight

Initial vertical velocity is zero.

Step 2 - Find the time of flight

Vertically: 45 = half times 9.81 times t squared.

Step 2 - Find the time of flight

t squared = 2(45)/9.81 = 9.17.

Step 2 - Find the time of flight

t = 3.03 s.

Step 3 - Find the horizontal range

No horizontal acceleration, so range = horizontal speed times time.

Step 3 - Find the horizontal range

R = 12 times 3.03 = 36.4 m.

Step 3 - Find the horizontal range

The projectile lands 36.4 m from the base of the cliff.

Step 4 - Find the vertical velocity at impact

Vertical velocity at impact: v_y = u_y + gt.

Step 4 - Find the vertical velocity at impact

v_y = 0 + 9.81(3.03) = 29.7 m/s downward.

Step 4 - Find the vertical velocity at impact

The horizontal velocity stays at 12 m/s throughout.

Step 5 - Calculate the resultant speed and angle at impact

Speed at impact = magnitude of the resultant velocity.

Step 5 - Calculate the resultant speed and angle at impact

v = square root of (12 squared + 29.7 squared) = square root of (144 + 882).

Step 5 - Calculate the resultant speed and angle at impact

v = 32.0 m/s.

Step 5 - Calculate the resultant speed and angle at impact

The angle below horizontal: arctan(29.7/12) = 68 degrees. The ball hits at 32.0 m/s, 68 degrees below horizontal.