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A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of 30 with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground ?

\left [ g=10 m/s^{2},sin30^{0}=1/2,cos30^{0}=\sqrt{3}/2 \right ]

Option: 1

5.20 m


Option: 2

4.33 m


Option: 3

2.60 m


Option: 4

8.66 m.


Answers (1)

best_answer

Height of building = 10m

The ball projected from the roof of the building will be back to the roof of a height of 10 m after covering the maximum horizontal range.

Maximum horizontal range  (R)= \frac{u^{2}\sin 2\Theta }{g}

or    R= \frac{\left ( 10^{2} \right )\times \sin 60^{\circ} }{10}= 10\times 0.866

or    R= 8.66

 

Posted by

Divya Prakash Singh

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