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?

Question

sidsiddhartha · Answer

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