NEET Class neet Answered
The ceiling of a hall is 40m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56m/s without hitting the ceiling of the hall is?
Asked by www.sukanyaarumughamk | 05 Aug, 2019, 11:07: AM
Expert Answer
Height h reached in projectile motion is given by, h = (u2 sin2α)/(2g) ....................(1)
where u is projection speed, α is projection angle and g i s acceleration due to gravity.
substituting the given values for u and h in eqn.(1),
we get projection angle α : sin2α = (2gh)/u2 = (2×9.8×40)/(56×56) or α = 30°
Horizontal range R for projctile motion : R = u2 sin2α /(2g) = (56×56×√3)/(2×2×9.8) = 138.56 m
Answered by Thiyagarajan K | 05 Aug, 2019, 12:19: PM
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