Question

A ball of mass .1kg is thrown at an angle 60 degrees with the horizontal with a speed 50 m/s. What is the kinetic energy at the highest point of the trajectory?

Answer 1

we know that at the highest point in the trajectory velocity in the Y direction is zero but it has a value in the X direction which is calculated by
\[Vx=(V0)*\cos \Theta\]
\[Vx=50*\cos60=25\]
now kinetic energy=\[KE=\frac{ 1 }{ 2 }*(mv ^{2})\]
ur m is given and v=25
now just calculate