Question

A projectile is fixed with Kinetic Energy of $1K$ Joule. If the range is maximum, what is the kinetic energy at the highest point.

Answer 1

The initial kinetic energy is given as,

$E=\frac{1}{2}m{v}^2$

$\frac{1}{2}m{v}^2=1000J$

Since for the maximum horizontal range the angle will be ${45}^{\circ }$.

The kinetic energy at the highest point is given as,

$E^{\prime }=\frac{1}{2}m{\left(v\cos \theta \right)}^2$

$=\frac{1}{2}m{v}^2{\cos }^2{45}^{\circ }$

$=1000\times \frac{1}{2}$

$=500J$

Thus, the kinetic energy at the highest point is $500J$.