Question

The horizontal range and maximum height of a projectile are R and H respectively. If a constant horizontal retardation a = g/4 is imparted to the projectile due to wind, then its new horizontal range will be:

• A. R - 2H
• B. R - H
• C. R - H/2
• D. R - 4H

Answer 1

Answer: (B)

R - H

Answer 2

The time of flight ($T$) is determined by the vertical motion and remains unchanged. The original horizontal range is $R = v_{0x} T$. The maximum height is $H = \frac{v_{0y}^2}{2g}$. A constant horizontal retardation $a = g/4$ means the horizontal acceleration $a_x = -g/4$. The new horizontal range $R'$ is given by $R' = v_{0x} T + \frac{1}{2} a_x T^2$. Substituting $a_x = -g/4$, we get $R' = R - \frac{1}{2} (\frac{g}{4}) T^2 = R - \frac{gT^2}{8}$. Using $T = \frac{2v_{0y}}{g}$, we find $\frac{gT^2}{8} = \frac{g}{8} (\frac{4v_{0y}^2}{g^2}) = \frac{v_{0y}^2}{2g}$. Since $H = \frac{v_{0y}^2}{2g}$, we have $\frac{gT^2}{8} = H$. Therefore, $R' = R - H$.