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 throwi - Physics MCQ AIEEE 2003 2026 | ExamTest.live

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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 3 0^{\circ} with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? (Take g = 10 m/s^{2})

Solution & Step-by-step Explanation

The ball is at a height of 10 m from the ground at two points: the point of projection and the point where it returns to the same horizontal level.The horizontal distance between these two points is the horizontal range R : R = \frac{u ^{2} sin 2 θ}{g} Given u = 10 m/s and θ = 3 0^{\circ} : R = \frac{1 0 ^{2} sin ( 2 \times 3 0 ^{\circ} )}{10} = \frac{100 \times sin 6 0 ^{\circ}}{10} = 10 \times \frac{3}{2} = 53 \approx 5 \times 1.732 = 8.66 m

Try it yourself before checking the explanation above.

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 3 0^{\circ} with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? (Take g = 10 m/s^{2})

5.20 m

4.33 m

2.60 m

8.66 m

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