Question

G = 10 N/kg or 10 m/s2

A tennis ball of mass 0.2 kg drops from a height of 20 m. When it hits the ground, it bounces to a height of 14 m.

Describe what happens next
What was the Potential Energy (EP) of the ball before it fell?
What was the efficiency of the bounce?
How fast will Danielle be moving as she reaches the bottom of the hill?

Answer 1

a)
`U_(g) = 40\,J`, b)
`\eta = 70\,\%`, c)
`v = 20\,(m)/(s)`

Step-by-step explanation:

a) The initial potential energy is:

`U_(g) = (0.2\,kg)\cdot \left(10\,(m)/(s^(2)) \right)\cdot (20\,m)`

`U_(g) = 40\,J`

b) The efficiency of the bounce is:

`\eta = \left((14\,m)/(20\,m) \right)* 100\,\%`

`\eta = 70\,\%`

c) The final speed of Danielle right before reaching the bottom of the hill is determined from the Principle of Energy Conservation:

`K = U_(g)`

`U_(g) = (1)/(2)\cdot m \cdot v^(2)`

`v = \sqrt{(2\cdot U_(g))/(m) }`

`v = \sqrt{(2\cdot (40\,J))/(0.2\,kg) }`

`v = 20\,(m)/(s)`