A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m-s at the angle of 30- with the horizontal- How far from the throwing point will be the ball be at the height of 10 m from the ground -Take- g-10 m-s2

Horizontal range is required R=u^2sin 2θg=(10)^2sin (2× 30^∘)10 ∴ R=5√(3)=8.66 m Hence, option (D) is correct. ...

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Standard XII

Physics

Application of Projectile to a Height

A boy playing...

Question

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

5.20 m

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4.33 m

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2.60 m

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8.66 m

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Solution

The correct option is D $8.66 m$

Horizontal range is required

$R = \frac{u^{2} sin 2 θ}{g} = \frac{(10)^{2} sin (2 \times 30^{\circ})}{10}$

$∴ R = 5 \sqrt{3} = 8.66 m$

Hence, option $(D)$ is correct.

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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 the angle of 30∘ with the horizontal. How far from the throwing point will be the ball be at the height of 10 m from the ground ? (Take, g=10 m/s2)

The correct option is D 8.66 m Horizontal range is required R=u2sin2θg=(10)2sin(2×30∘)10 ∴R=5√3=8.66 m Hence, option (D) is correct.

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

The correct option is D $8.66 m$

Horizontal range is required

$R = \frac{u^{2} sin 2 θ}{g} = \frac{(10)^{2} sin (2 \times 30^{\circ})}{10}$

$∴ R = 5 \sqrt{3} = 8.66 m$

Hence, option $(D)$ is correct.