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(D) The ball is thrown from a height of $10 \, m$ and we need to find the horizontal distance when it returns to the same height of $10 \, m$. This is

Vedclass · Accepted Answer

$8.66$

Vedclass · Answer

(D) The ball is thrown from a height of $10 \, m$ and we need to find the horizontal distance when it returns to the same height of $10 \, m$. This is equivalent to the horizontal range of a projectile launched from the ground. The formula for the horizontal range $R$ is given by $R = \frac{u^2 \sin(2	heta)}{g}$. Given: $u = 10 \, m/s$,$	heta = 30^{\circ}$,$g = 10 \, m/s^2$. Substituting the values: $R = \frac{10^2 	imes \sin(2 	imes 30^{\circ})}{10} = \frac{100 	imes \sin(60^{\circ})}{10} = 10 	imes \frac{\sqrt{3}}{2}$. $R = 5 	imes 1.732 = 8.66 \, m$. Thus,the ball will be at the height of $10 \, m$ at a distance of $8.66 \, m$ from the throwing point.

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