The horizontal range and maximum height attai | vedclass

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Question

(D) The vertical motion of the projectile is independent of the horizontal motion. Since the vertical acceleration remains $-g$,the maximum height $H$

Vedclass · Accepted Answer

$(R+H), H$

Vedclass · Answer

(D) The vertical motion of the projectile is independent of the horizontal motion. Since the vertical acceleration remains $-g$,the maximum height $H$ remains unchanged.For the horizontal range,the initial range is $R = u_x T$,where $T = \frac{2 u_y}{g}$ is the time of flight.The new horizontal range $R^{\prime}$ with horizontal acceleration $a_x = \frac{g}{4}$ is given by:$R^{\prime} = u_x T + \frac{1}{2} a_x T^2$Substituting $R = u_x T$ and $T^2 = \frac{4 u_y^2}{g^2}$:$R^{\prime} = R + \frac{1}{2} \left(\frac{g}{4}\right) \left(\frac{4 u_y^2}{g^2}\right)$Since $H = \frac{u_y^2}{2g}$,we have $u_y^2 = 2gH$. Substituting this:$R^{\prime} = R + \frac{g}{8} \left(\frac{4(2gH)}{g^2}\right) = R + \frac{g}{8} \left(\frac{8gH}{g^2}\right) = R + H$Thus,the new range is $(R+H)$ and the height is $H$.

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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