Question

an object weighting 100g is thrown upwards from the ground at a speed of 100 m/s. where will the potential energy of the object be two thirds of the kinetic energy?

Answer 1

Explanation:

so angle beteween the initial velocity and

ground is 90°

we know by the equations of projectile motion

H=u²sin²ø/2g//,let g=10m/s² or you can also use

V²=u²-2gs (negative since upward direction)

at heighest point final velocity =0m/s

0²=u²-2gs

s=u²/2g //

s=100²/2×10

s=10000/20

S=500m

K.E=1/2mv²

v=100m/s,m=100g=1/10Kg

K.E=1/2×1/10×10000=500J

According to the question .....

when will potential energy be 2/3rd of kinetic

energy...... so we need to find at what height

2/3×K.E=P.E

2×500/3=1/10×10×h

h=333.33m