A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle 30∘ with horizontal. How far from the throwing point will the ball be at the height of 10m from the ground (g=10m/s2)

R=u2sin2θg=10×10×sin60o10=53=8.66m

A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle 30∘ with horizontal. How far from the throwing point will the ball be at the height of 10m from the ground (g=10m/s2)

R=u2sin2θg=10×10×sin60o10=53=8.66m

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A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle $30^{\circ}$ with horizontal. How far from the throwing point will the ball be at the height of 10m from the ground $(g = 10 m / s^{2})$

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8.66 m

5.20 m

4.33 m

2.60 m

answer is D.

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Detailed Solution

$R = \frac{u^{2} \sin 2 θ}{g}$

$= \frac{10 \times 10 \times \sin 60^{o}}{10}$

$= 5 \sqrt{3} = 8.66 m$

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A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle 30∘ with horizontal. How far from the throwing point will the ball be at the height of 10m from the ground (g=10m/s2)

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A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle $30^{\circ}$ with horizontal. How far from the throwing point will the ball be at the height of 10m from the ground $(g = 10 m / s^{2})$