A ball is launched with an initial velocity of 28.5 m/s at a 45° angle from the top of a cliff that is 10.0 m above the water below. Use g = 9.80 m/s2 to simplify the calculations. Note: You could answer these questions using projectile motion methods, but try using an energy conservation approach instead.

(a) What is the ball's speed when it hits the water?

m/s

(b) What is the ball's speed when it reaches its maximum height?

m/s

(c) What is the maximum height (measured from the water) reached by the ball in its flight?

m

2.

A boy reaches out of a window and tosses a ball straight up with a speed of 15 m/s. The ball is 30 m above the ground as he releases it. Use energy to find the following.

(a) What is the ball's maximum height above the ground?

m

(b) What is the ball's speed as it passes the window on its way down?

m/s

(c) What is the speed of impact on the ground?

Question

Physics

Tanya Briggs

18 July, 04:17

A ball is launched with an initial velocity of 28.5 m/s at a 45° angle from the top of a cliff that is 10.0 m above the water below. Use g = 9.80 m/s2 to simplify the calculations. Note: You could answer these questions using projectile motion methods, but try using an energy conservation approach instead.

(a) What is the ball's speed when it hits the water?

m/s

(b) What is the ball's speed when it reaches its maximum height?

m/s

(c) What is the maximum height (measured from the water) reached by the ball in its flight?

m

2.

A boy reaches out of a window and tosses a ball straight up with a speed of 15 m/s. The ball is 30 m above the ground as he releases it. Use energy to find the following.

(a) What is the ball's maximum height above the ground?

m

(b) What is the ball's speed as it passes the window on its way down?

m/s

(c) What is the speed of impact on the ground?

+1

Answers (1)

Know the Answer?

Lyric Mccoy · Answer 1

When it hits the water, the initial kinetic energy will increase by additional potential energy of mgh where h is 10 m.

Initial kinetic energy

= 1/2 x m x (28.5) ²

= 406.125 m

Extra potential energy added

= m x 9.8 x 10

= 98m

Total energy

= 504.125 m J

If v the the velocity at the bottom

1/2 m v² = 504.125 m

v = 31.75 m/s

b) The velocity at the top will be equal to the horizontal component of initial velocity

= 28.5 cos 45

= 20.15 m / s

c) kinetic energy at the top

= 1/2 m x (20.15) ²

= 203 m

Loss of energy at the top

= 406.125 m - 203 m

= 203.125 m

This energy has been converted into potential energy

mgH = 203 m

H = 203 / 9.8

= 20.71 m