Complete Question:
A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level
a) Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.
b) What is the height, hn in meters?
Answer:
a) Energy = mghₙ
b) Height, hₙ = 5.02 m
Explanation:
a) Total energy in terms of maximum height
Let maximum height be hₙ
At maximum height, velocity, V=0
Total mechanical energy , E = mgh + 1/2 mV^2
Since V=0 at maximum height, the total energy in terms of maximum height becomes
Energy = mghₙ
b) Height, hₙ in meters
mghₙ = mgh + 1/2 mV^2
mghₙ = m(gh + 1/2 V^2)
Divide both sides by mg
hₙ = h + 0.5 (V^2)/g
h = 2.15m
g = 9.8 m/s^2
V = 7.5 m/s
hₙ = 2.15 + 0.5(7.5^2)/9.8
hₙ = 2.15 + 2.87
hₙ = 5.02 m
