A projectile is fired at 53° with speed u. Take sin53°=4/5, cos53°=3/5 and tan37°= $\frac{3}{4}$ . The time after projection when its velocity first makes an angle 37° above the horizontal is:
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A projectile is fired at 53° with speed u. Take sin53°=4/5, cos53°=3/5 and tan37°=3/4. The time after projection when its velocity first makes an angle 37° above the horizontal is:

G1 Physics · Accepted Answer

Horizontal velocity is vₓ = 3u/5. At the required instant, vᵧ/vₓ = tan37° = 3/4. Hence vᵧ = (3/4)(3u/5) = 9u/20. But vᵧ = 4u/5 − gt = 16u/20 − gt. Therefore gt = 7u/20 ⇒ t = 7u/(20g).