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

G1 Physics · Accepted Answer

Horizontal velocity is vₓ=3u/5. For 37° below the horizontal, vᵧ/vₓ = −3/4. Thus vᵧ = −9u/20. Now 4u/5 − gt = −9u/20 ⇒ 16u/20 + 9u/20 = gt ⇒ t = 25u/(20g) = 5u/(4g).