work and energy

Question

A basketball player throws a basketball m = 1 kg straight up with an initial speed of v₀ = 5.5 m/s. The ball leaves his hand at shoulder height h₀ = 2.1 m. Let gravitational potential energy be zero at ground level.

a) give the total mechanical energy of the ball E in terms of the maximal height hm it reaches, the mass m, and the gravitational acceleration g. E=

b)what is the height hm in meters?

Answer 1

When the ball leaves the player's hands it has a certain amount of energy which is composed of 1) gravitational potential energy (E_GP) and 2) kinetic energy (E_K):

E = E_GP + E_K

Gravitational potential energy is calculated as: E_GP = mgh

Kinetic energy is calculated as: E_K = 1/2mv²

So: E = mgh + 1/2mv²

We are told m = 1kg, v = 5.5 m/s, h = 2.1 m so:

E = (1)(9.8)(2.1) + 1/2(1)(5.5²) = 35.705 joules

Energy is conserved if there is no friction (which we will assume) so the energy at the top of the ball's flight will also equal 35.705 joules. At that point, the velocity of the ball will reach zero just at the instant the ball changes direction to come back down. At that point, there will be no kinetic energy (since v = 0) so all the energy will be gravitational potential energy. Therefore:

E = mgh or

E = (1)(9.8)(h)

E = 9.8h

Since E = 35.705 then:

35.705 = 9.8h

h = 35.705/9.8

h = 3.6 meters