Question

A rock of mass 200kg is dropped from a height of 200m. What is the potential energy at:
a) 0 seconds
b) 2 seconds
c) 4 seconds
d) just before it hits the ground

Answer 1

a) p= mgh
= 200*200*10= 400000 J
b) distance in 2 sec = g * t²/ 2
= 10 * 2² / 2
= 10 * 4 / 2
= 10 * 2
= 20 m
now, h= 200-20= 180 m

therefore, p = mgh = 200*10*180=360000J

again,
c) distance in 4 second = g * t²/ 2
= 10 * 4² / 2
= 10 * 16 / 2
=10 * 8
= 80m

now, h= 200-80= 120 m

therefore, p = mgh = 200*10*120=240000J

d) just before it hits the ground
p=o

Answer 2

a)
at 0 sec. h = 200 m
P.E = Mgh = 200× 9.8 × 200
= 392000 kg.m^2/s^2

b) ar t = 2sec

t = ✓2d/g

2^2 = 2×h/9.8

=> d = 19.6 m
hence h = 200-19.6 = 180.4 m

hence P.E = mgh = 200× 9.8 × 180.4
= 353584 J
c)
t = 4 sec
d = g•t^2/2 = 9.8×16/2 = 78.4 m
hence h = 200-78.4 = 121.6 m

therefore
P.E = Mgh = 200× 9.8× 121.6
= 238336 J

d) jut before it hits the ground
h = zero
hence P.E = 0 J