Q: A body is projected with velocity u at an angle of projection θ with the horizontal. The body makes 30° with horizontal at t = 2 second and then after 1 second it reaches the maximum height. Then find

(a) Angle of projection

(b) Speed of projection

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Sol. During the projectile motion, angle at any instant t is

$\displaystyle tan\alpha = \frac{v_y}{v_x} = \frac{usin\theta – g t}{u cos\theta}$

For t = 2 seconds, α =30°

$\Large \frac{1}{\sqrt{3}} = \frac{usin\theta – 2g}{u cos\theta}$ ….(i)

For t = 3 seconds, at the highest point α = 0°

$\Large 0 = \frac{usin\theta – 3g}{u cos\theta}$ ….(ii)

Solving (i) & (ii)

θ = 60° , u = 20√3 m/s