The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball is thrown with a speed of 40 $$ms^{-1}$$ can go without hitting the ceiling of the hall?

Asked by Abhisek | 1 year ago |  70

##### Solution :-

Speed of the ball, u = 40 ms-1

Maximum height, h = 25 m

In projectile motion, the maximum height achieved by a body projected at an angle $$\theta$$, is given as:

$$h = \dfrac{u^{2}\sin ^{2}\theta }{2g}25$$

$$= \dfrac{\left (40 \right )^{2}\sin ^{2}\theta }{2 \times 9.8}\sin ^{2}\theta$$

$$= 0.30625\sin \theta = 0.5534\theta = \sin ^{-1}\left (0.5534 \right ) = 33.60°$$

Horizontal range, R:

=$$\dfrac{u^{2}\sin 2\theta}{g}$$

$$\dfrac{\left (40 \right )^{2}\sin 2 \times 33.60}{9.8}$$

=$$\dfrac{1600 \times \sin 67.2}{9.8}$$

= $$\dfrac{1600 \times 0.922}{9.8}$$= 150.53 m

Answered by Pragya Singh | 1 year ago

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