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At what angle should a body be projected with a velocity 24 ms -1  just to pass over the obstacle 16 m high at a horizontal distance of 32 m . Assume g = 10 ms -1 ? Solution: what angle should If point of projection is taken as the origin of the coordinate system, the projected body must pass through a point having coordinates (32 m, 16 m) as the following figure. Suppose u be the initial velocity of the projectile. If θ be the angle of projection, then Horizontal component of initial velocity u x  = u cos θ. And vertical component of initial velocity u y  = u sin θ. Suppose the body passes through point P at time t. then horizontal distance covered is x = 32 m i.e., 32 = x = (u cos θ ) t or 32 = (24 cos θ) t …….(1) Similarly, vertical distance covered is y = 16 m i.e., 16 = y = (u sin θ)t-1/2gt 2 Or  16 = (24 sin θ)t – 1/2x 10x t 2  …..(2) From equation (1) we get : t = 32/(24 cos θ) Putting this value of t in equati...