Question

a body is projected at an angle such that K.E. at the highest point is reduced to half the energy at point of projection. find angle of projection

Answer 1

Let a body of mass m is projected at an angle Φ with intial speed u.
at initial position , only kinetic energy exists [because there is no height to gain potential energy ]
hence, initial energy = 1/2 mu² ----(1)
at highest position, exists only horizontal component of velocity e.g., $v_x$=ucosΦ
hence, kinetic energy = 1/2 m(ucosΦ)² ---(2)

A/C to question,
kinetic energy at highest point = 1/2 × energy at point of projection.
=> 1/2 m(ucosΦ)² = 1/2 × 1/2 mu²
=> cos²Φ = 1/2
=> cosΦ = ± 1/2
hence, Φ = 45° or 135°
but Φ < 90° , so, Φ ≠ 135°
hence, Φ = 45°