A particle is fired with velocity `u` making angle `theta` with the horizontal.What is the change in velocity when it is at the highest point?

Options

A. `1/2 mu^(2) sin^(2)theta`
B. `1/2 mu^(2)cos^(2)theta`
C. `1/2 mu^(2)`
D. `1/2 mu^(2) sin^(2) 2theta`

Answer

A

Explanation

The vertical component of velocity is reduced to zero. The corresponding kinetic energy is converted into potential energy (law of conservation of energy)
`therefore PE = 1/2 m(u sin theta)^(2) = 1/2 mu^(2) sin^(2)theta`