TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

The motion of projectile is as shown in figure,Let initial velocity of projection is u, so initial kinetic energy is
KE1=‌

1

2

mu2...(i)
At highest point, the vertical component of velocity is zero, while the horizontal component is u‌cos‌θ, which is non-zero. So, the kinetic energy at this point is
KE2=‌

1

2

m(u‌cos‌θ)2=‌

1

2

mu2cos2θ
=KE1cos260∘=‌

K

4

(∵KE1=K and cos‌60∘=1/2)