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

1 Answer

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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