Question

If the kinetic energy of an oblique projectile at its
maximum height is half of its initial kinetic energy
then the angle of throw with the vertical is
(1) 30°
(2) 37°
(3) 45°
(4) 60°​

Answer 1

answer : option (3) 45°

explanation : Let a body is projected with speed u at an angle α with horizontal.

so, initial kinetic energy, Ki = 1/2 mu² ...(2)

then, velocity at maximum height , v = ucosα

now, kinetic energy at maximum height, K = 1/2 mv²

K = 1/2 m(ucosα)²

K = 1/2 mu²cos²α ....(2)

a/c to question,

kinetic energy at maximum height = 1/2 × initial kinetic energy

or, 1/2 mu²cos²α = 1/2 × 1/2 mu²

or, cos²α = 1/2 = (1/√2)²

or, cosα = cos45° => α = 45°

hence, body is projected at angle 45° with horizontal.