A boy throws a ball from shoulder height at an initial velocity of 30 m/s. Spending 4.8 sin air, the ball
is caught by another boy as the same shoulder-height level. What is the angle of projection?
(A) 37°
(B) 30°
(C) 53°
(D) 60°
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Hey dear,
● Answer -
θ = 51.63°
◆ Explaination -
# Given -
u = 30 m/s
T = 4.8 s
# Solution -
Time of flight of the ball thrown in air is calculated by formula -
T = 2u.sinθ/g
4.8 = 2 × 30 × sinθ / 9.8
sinθ = (4.8×9.8) / (2 × 30)
sinθ = 0.784
θ = arcsin(0.784)
θ = 51.63°
Therefore, most appropriate answer would be 53°.
Thanks for asking...
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