The horizontal range and maximum height attai | vedclass

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Question

(d)

$H=H$ (as vertical component of acceleration has not changed)

$R^{\prime} =u_x T+\frac{1}{2} a_x T^2$

$=R+\frac{1}{2} 	imes \frac{g}{4}

Vedclass · Accepted Answer

$(R+H), H$

Vedclass · Answer

(d)

$H=H$ (as vertical component of acceleration has not changed)

$R^{\prime} =u_x T+\frac{1}{2} a_x T^2$

$=R+\frac{1}{2} 	imes \frac{g}{4} 	imes \frac{4 u^2 \sin ^2 	heta}{g^2}$

$=R \frac{u^2 \sin ^2 	heta}{2 g}=(R+H)$

The horizontal range and maximum height attained by a projectile are $R$ and $H$, respectively. If a constant horizontal acceleration $a=g / 4$ is imparted to the projectile due to wind, then its horizontal range and maximum height will be

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