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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
answered-by-expert 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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