Question

a body is projected with velocity U such that its horizontal range and maximum vertical Heights are same the maximum height is​

Answer 1

Hey Dear,

◆ Answer

Hmax = u².sin²76° / 2g

◆ Explanation -

Maximum height achieved by projectile is given as -

Hmax = u².sin²θ / 2g

Range of projectile is given as -

Rmax = u².sin2θ / g

Given that,

Hmax = Rmax

u².sin²θ / 2g = u².sin2θ / g

sinθ.sinθ / 2 = 2 sinθ.cosθ

tanθ = 4

θ = 75.96° ~ 76°

So now maximum height is given as -

Hmax = u².sin²θ / 2g

Hmax = u².sin²76° / 2g

Thanks dear...