If a stone is to hit a point which is at a ho | vedclass

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(B) The equations of motion for a projectile are:Horizontal displacement: $d = (u \cos 	heta) t$Vertical displacement: $h = (u \sin 	heta) t - \frac

Vedclass · Accepted Answer

$\frac{d}{\cos 	heta} \sqrt{\frac{g}{2(d 	an 	heta - h)}}$

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(B) The equations of motion for a projectile are:Horizontal displacement: $d = (u \cos 	heta) t$Vertical displacement: $h = (u \sin 	heta) t - \frac{1}{2} g t^2$From the horizontal equation,we find the time of flight to reach distance $d$:$t = \frac{d}{u \cos 	heta}$Substitute this expression for $t$ into the vertical displacement equation:$h = (u \sin 	heta) \left( \frac{d}{u \cos 	heta} ight) - \frac{1}{2} g \left( \frac{d}{u \cos 	heta} ight)^2$$h = d 	an 	heta - \frac{g d^2}{2 u^2 \cos^2 	heta}$Rearranging to solve for $u^2$:$\frac{g d^2}{2 u^2 \cos^2 	heta} = d 	an 	heta - h$$u^2 = \frac{g d^2}{2 \cos^2 	heta (d 	an 	heta - h)}$Taking the square root of both sides:$u = \frac{d}{\cos 	heta} \sqrt{\frac{g}{2(d 	an 	heta - h)}}$

If a stone is to hit a point which is at a horizontal distance $d$ and at a vertical height $h$ above the point from where the stone is launched,what is the initial speed $u$ if the stone is launched at an angle $\theta$?

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