Question

A basketball is shot from a height of 2 m at a speed of 4.9 m/s. How fast is it going as it reaches the height of the rim, which is 10 ft high? Answer in m/s. Hint: try using energy rather than freefall acceleration.

Answer 1

The total energy of the basketball when it is shot
E = KE + PE
The kinetic energy of the ball is
KE = (1/2) m v²
KE = (1/2) m x (4.9 m/s)²
KE =  12.005 x m
The potential energy at the height of 2 m.
P E = m g h
PE = m x 9.8 m/s² x 2 m
PE = 19.6 x m
The total energy at the height 10 ft.
E = PE + K E
The PE = m g h = m x 9.8 x (10 x 0.3048 m)
PE =  29.8704 m
The kinetic energy is
KE = (1/2) m v_f²
We need to find this velocity.
Equating the total energies at this two situations.
12.005 x m + 19.6 x m =  29.8704 m + (1/2) m v_f²
12.005 + 19.6 =  29.8704 + (1/2) v_f²
v_f² = 3.4692
v_f = 1.86 m/s