A ball is thrown at 30^{circ} with the horizo | vedclass

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(B) The ball is projected from a height of $20 \,m$ and we need to find the horizontal distance where it returns to the same vertical level ($20 \,m$

Vedclass · Accepted Answer

$14.6$

Vedclass · Answer

(B) The ball is projected from a height of $20 \,m$ and we need to find the horizontal distance where it returns to the same vertical level ($20 \,m$ from the ground).This is equivalent to finding the horizontal range of a projectile launched at an angle $	heta$ on a level plane.The formula for the horizontal range $R$ is given by:$R = \frac{u^2 \sin(2	heta)}{g}$Given:Initial speed $u = 13 \,ms^{-1}$Angle of projection $	heta = 30^{\circ}$Acceleration due to gravity $g = 10 \,ms^{-2}$Substituting the values:$R = \frac{(13)^2 	imes \sin(2 	imes 30^{\circ})}{10}$$R = \frac{169 	imes \sin(60^{\circ})}{10}$$R = \frac{169 	imes \frac{\sqrt{3}}{2}}{10}$$R = \frac{169 	imes 1.732}{20}$$R = \frac{292.708}{20} \approx 14.635 \,m$Rounding to one decimal place, we get $R = 14.6 \,m$.

$A$ ball is thrown at $30^{\circ}$ with the horizontal from the top of a roof $20 \,m$ high with a speed of $13 \,ms^{-1}$. At what distance from the throwing point will the ball, once again, be at a height of $20 \,m$ from the ground (in $\,m$)? $(g = 10 \,ms^{-2})$

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