Question

At t = 0 a 1.2 kg ball is thrown from the top of a tall tower with velocity v= (11 m/s)i + (29 m/s)j . What is ΔU of the ball-Earth system between t = 0 and t = 4.3 s (still free fall)?

Answer 1

Solution:

Given:

mass of the ball (m) = 1.2 kg
initial velocity vector (v) = (11 m/s) i + (29 m/s) j

V_0y = 29 m/s

and t = 4.3 s
from equation of motion due to acceleration,
h = V_0y t - (1/2) g t² = (29 m/s)(4.3 s) - (0.5)(9.8 m/s²)(4.3 s)² = 34.099 m

Therefore: change in potential energy
U = m g h = (1.2 kg)(9.8 m/s²)( 34.099 m) = 401.4 J