Question

a body is projected such that it clears a wall of height 24 m which is at a distance of 32 m from the point of projection. Then the least required velocity is u m/s, so u^2 is​

Answer 1

Dear Student,

● Answer -

u² = 680 m²/s²

◆ Explanation -

Considering the body moving in projectile motion,

Hmax = 24 m

R = 32×2 = 64 m

Horizontal range can be given as -

R = u²sin2θ/g

64 = u²sin2θ/g

u²/g = 64/sin2θ ...(1)

Maximum height is given by -

Hmax = u²sin²θ/2g

24 = u²sin²θ/2g

u²/g = 48/sin²θ ...(2)

From (1) & (2),

64/sin2θ = 48/sin²θ

48 × 2sinθ.cosθ = 64 × sinθ.sinθ

sinθ/cosθ = 48 × 2 / 64

tanθ = 1.5

θ = arctan(1.5)

θ = 56.31

Putting this in (2),

u²/g = 48/sin²θ

u² = 48 × 9.8 / sin²56.31

u² = 680 m²/s²

Hope I am useful...