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Motion in a Plane

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Question : 15 of 32
Marks: +1, -0
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−1^{-1} can go without hitting the ceiling of the hall?
Solution:  
Here, H = 25 m; u = 40 m s −1^{-1}
Suppose that the ball is thrown at an angle q with the horizontal.
We know, H=u2sin⁡2θ2gH= \frac{u^{2} \sin^{2} \theta}{2g}
∴25=(40)2sin⁡2θ2×9.8\therefore 25= \frac{(40)^{2} \sin^{2} \theta}{2 \times 9.8}
or sin⁡2θ=25×2×9.8(40)2\sin^{2} \theta = \frac{25 \times 2 \times 9.8}{(40)^{2}} or sin⁡θ=49040=0.5534\sin \theta = \frac{\sqrt{490}}{40}=0.5534 or θ=33.6∘\theta = 33.6^{\circ}
Now, R=u2sin⁡2θg=(40)2sin⁡2(33.6∘)9.8R= \frac{u^{2} \sin 2\theta}{g}= \frac{(40)^{2} \sin 2(33.6^{\circ})}{9.8}
=(40)2sin⁡67.2∘9.8= \frac{(40)^{2} \sin 67.2^{\circ}}{9.8} =(40)2×0.92199.8=150.5 m= \frac{(40)^{2} \times 0.9219}{9.8}=150.5\ \text{m}
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