Question

A 0.690 kg snowball is fired from a cliff 8.25 m high with an initial velocity of 10.3 m/s, directed 32.0° above the horizontal. (a) Using energy techniques, find the speed of the snowball as it reaches the ground below the cliff. What is that speed (b) if the launch angle is changed to 32.0° below the horizontal and (c) if the mass is changed to 1.40 kg?

Answer 1

a) 16.4 m/s

b) 16.4 m/s

c) 16.4 m/s

Step-by-step explanation:

a)

m = mass of the snowball = 0.690 kg

h = height of the cliff = 8.25 m

v₀ = initial speed of ball at the time of launch = 10.3 m/s

v = speed of the ball as it reach the ground

Using conservation of energy

initial kinetic energy + initial potential energy at the cliff = final kinetic energy just before reaching the ground

(0.5) m v₀² + mgh = (0.5) m v²

(0.5) v₀² + gh = (0.5) v²

(0.5) (10.3)² + (9.8 x 8.25) = (0.5) v²

v = 16.4 m/s

b)

As the launch angle is changed, the speed of the ball just before reaching the ground remain the same as the final speed does not depend on the angle of launch.

v = 16.4 m/s

c)

As the mass is changed, the speed of the ball just before reaching the ground remain the same as the final speed does not depend on the mass of the ball.

v = 16.4 m/s