A boy playing on the roof of a 10, m high bui | vedclass

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Question

(D) The boy throws the ball from the roof of a building of height $10\, m$. We need to find the horizontal distance from the throwing point when the b

Vedclass · Accepted Answer

$5\sqrt{3}$

Vedclass · Answer

(D) The boy throws the ball from the roof of a building of height $10\, m$. We need to find the horizontal distance from the throwing point when the ball is at a height of $10\, m$ from the ground. Since the ball is thrown from a height of $10\, m$ and we are looking for the distance when it is again at a height of $10\, m$,this corresponds to the horizontal range of the projectile on a horizontal plane passing through the point of projection. The formula for the horizontal range $R$ is given by: $R = \frac{u^2 \sin(2	heta)}{g}$ Given: Initial speed $u = 10\, m/s$ Angle of projection $	heta = 30^o$ Acceleration due to gravity $g = 10\, m/s^2$ Substituting the values: $R = \frac{(10)^2 \sin(2 	imes 30^o)}{10} = \frac{100 	imes \sin(60^o)}{10} = 10 	imes \frac{\sqrt{3}}{2} = 5\sqrt{3}\, m$ Thus,the ball will be at a distance of $5\sqrt{3}\, m$ from the throwing point.

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