PE = 450 m
Further explanation
Given
The initial problem might be like this
A 1.0-kg ball is thrown into the air with an initial velocity of 30 m/s.
Required
Potential energy
Solution
h max = vo²sin²θ/2g
h max = 30²sin²90/2.10
h max = 45 m
PE = m.g.h
PE = 1 x 10 x 45
PE = 450 m