The speed of a projectile at its highest point is v1 and at the point half the maximum height is v2.  If v1v2=25,then find the angle of projection

v1=v2cos⁡β=ucos⁡θ 0=v2sin⁡β2−2g(H/2) ⇒ v12=v22−gHAlso v1v2=cosβ=25From above v1=2gH3, v2=5gH3H=u2sin2⁡θ2g⇒sin2⁡θ=2gHcos2⁡θv12or tan2⁡θ=2gHv12  or  tan2⁡θ=2gH×32gHor tan⁡θ=3  or  θ=60∘