Question

A body is projected with a velocity of 40 ms^-1. After 2 s it crosses a vertical pole of height 20.4 m. Find the angle of projection and horizontal range of projectile, (g = 9.8 ms^-2).

Answer 1

Given: u = 40 ms^-1, t = 2 s, y = 20.4 m,

a_y = -9.8 m/s²

To find: 

(i) Angle of projection (θ)

(ii) Horizontal range of projectile (R)

Formulae: (i) y = u_y t + \(\frac{1}{2}\) a_y t²

(ii) R = \(\frac{u^2\,sin\,2\theta}{g}\)

Calculation: Taking vertical upward motion of the projectile from point of projection up to the top of vertical pole we have

u_y = 40 sinθ,

From formula (i),

∴ 20.4 = 40 sinθ × 2 + \(\frac{1}{2}\) (-9.8) × 2²

∴ 20.4 = 80 sinθ – 19.6

or sinθ = \(\frac{(20.4+19.6)}{80}=\frac{1}{2}\)

or θ = 30°.

From formula (ii),

Horizontal range = \(\frac{40^2}{9.8}\) sin 2 × 30°

= 141.4 m

The angle of projection is 30°. The horizontal range of projection is 141.4 m