Question : The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m/s can go without hitting the ceiling of the hall ?
Doubt by Jaskirat
Let θ be the angle of projection.
Solution :
Hmax = 25 m
u = 40 m/s
Let θ be the angle of projection.
Hmax=u²sin²θ/2g
25=(40)²sin²θ/(2×9.8) [g=9.8 m/s²]
25=(40)²sin²θ/(2×9.8) [g=9.8 m/s²]
25=1600sin²θ/(2×9.8)
(25×2×9.8)/1600 = sin²θ
√[(50×9.8)/1600] = sinθ
sinθ =√[490/1600]
sinθ = √[49/160]
sinθ = √(0.30625)
(25×2×9.8)/1600 = sin²θ
√[(50×9.8)/1600] = sinθ
sinθ =√[490/1600]
sinθ = √[49/160]
sinθ = √(0.30625)
sinθ = 0.553
cosθ = √[1-sin²θ]
cosθ = √[1-(0.553)²]
cosθ = √[1-0.306]
cosθ = √[1-0.306]
cosθ = √[0.694]
cosθ = 0.833
Maximum Horizontal Distance (Range)
R = u²sin2θ/g
R = u²(2sinθcosθ)/g
R = 2u²sinθcosθ/g
R = [2(40)²×0.553×0.833]/9.8
R = [3200×0.461]/9.8
R = [1475.2]/9.8
R = 150.53 m
R = u²sin2θ/g
R = u²(2sinθcosθ)/g
R = 2u²sinθcosθ/g
R = [2(40)²×0.553×0.833]/9.8
R = [3200×0.461]/9.8
R = [1475.2]/9.8
R = 150.53 m