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Ceiling of hall is 40m high. For maximum horizontal distance, angle at which ball may be thrown with speed of 56 m/s without hitting ceiling of hall

Tinsel , 8 Years ago
Grade 12
anser 1 Answers
Arun
V² sin² Ф -  2 g H = 0²    =>  V² sin² Ф = 2 g H  => 56 * 56 sin² Ф = 2 * 9.8 * 40
         Sin² Ф = 1/4  => sin Ф = 1/2  =>  30 degrees
Maximum height = H        Range = R    angle of projection = Ф
 H / R  =  (1/4 ) tan Ф 
 H = 40 m  
   tan Ф = 1/√3
Maximum range = R = 160 √3 m at this angle 30 deg
Last Activity: 8 Years ago
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