Question

A body is projected at an angle 60 with the horizontal with kinetic energy k the kinetic energy of the body at the highest point will be

Answer 1

Let the initial velocity be v. The projectile starts with [math]60^{\circ}[/math]

Now the horizontal component of velocity is same throughout the projectile.

Horizontal component [math]v_{horizontal} = v*\cos{60^{\circ}}[/math]

At the point at which the projectile becomes [math]30^{\circ}[/math]

[math]\cos{30^{\circ}} = \frac{v_{horizontal}}{v}[/math]

or [math]v = \frac{v_{horizontal}}{\cos{30^{\circ}}}[/math]

which gives us [math]v = 0.577*v[/math]

Kinetic energy at beginning [math]KE = \frac{1}{2}*m*v^2[/math]

Kinetic energy when projectile at [math]30^{\circ}[/math] = [math]KE_{30^{\circ}} = \frac{1}{2}*m*(0.577*v)^2[/math]

Therefore kinetic energy when projectile at [math]30^{\circ}[/math] becomes [math]0.3333*KE[/math]