A projectile is thrown such that it achieves | vedclass

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(C) For maximum range,the angle of projection is $	heta = 45^\circ$.At the highest point of the trajectory,the vertical component of velocity becomes

Vedclass · Accepted Answer

$(R/2, R/4)$

Vedclass · Answer

(C) For maximum range,the angle of projection is $	heta = 45^\circ$.At the highest point of the trajectory,the vertical component of velocity becomes zero,and the velocity is minimum,equal to $u \cos 	heta$.The coordinates of the highest point are $(x, y) = (R/2, H)$,where $R$ is the range and $H$ is the maximum height.The formula for maximum height is $H = \frac{u^2 \sin^2 	heta}{2g}$ and for range is $R = \frac{u^2 \sin 2	heta}{g}$.For $	heta = 45^\circ$,$H = \frac{u^2 (1/\sqrt{2})^2}{2g} = \frac{u^2}{4g}$ and $R = \frac{u^2 \sin 90^\circ}{g} = \frac{u^2}{g}$.Thus,$H = R/4$.The coordinates of the highest point are $(R/2, R/4)$.

$A$ projectile is thrown such that it achieves a maximum range $R$. What are the coordinates of the point where its velocity is minimum?

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