Question

A body is projected with velocity of 40 metre per sec. After 2 second it crosses a vertical pole of height 20.4 m. Calculate the angle of projection and horizontal range

Answer 1

y = u Sin theta* t - g t^2 /2
20.4 = 40* Sin theta* 2 - 10*4/2
=> sin theta = 0.505 ie theta = 30 deg approximately.

Range = u^2 Sine 2*theta/ g
= 80 sqrt3 m approximately

Answer 2

vertical component of initial velocity = 40sinФ
    use formula, 
      y = 40sinФ.t - 1/2gt²
here,
  t = 2 sec 
  y = 20,4 m 
  g = 9.8 m/s²

20.4 = 80sinФ - 1/2 × 9.8 × 4 
40 = 80 sinФ
sinФ = 1/2 
Ф = 30°


now ,
   Range = u²sin2Ф/g
              = (40)² × √3/2/10      (9.8 ≈ 10 m/s² )
              =80√3 m