Question

The horizontal range and maximum height of a projectile are $R$ and $H$ respectively. If a constant horizontal acceleration $a=g/2$ is imparted to the projectile due to wind, then its new horizontal range will be:

• A. R+H
• B. R+2H
• C. R+H/2
• D. R

Answer 1

Answer: (B)

R+2H

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 $H = \frac{v_{0y}^2}{2g}$. The new horizontal range $R'$ is given by $R' = v_{0x} T + \frac{1}{2} a T^2$. Substituting $a = g/2$, we get $R' = R + \frac{1}{2} (\frac{g}{2}) T^2 = R + \frac{g}{4} T^2$. Using $T = \frac{2v_{0y}}{g}$, we find $R' = R + \frac{g}{4} (\frac{2v_{0y}}{g})^2 = R + \frac{v_{0y}^2}{g}$. Since $H = \frac{v_{0y}^2}{2g}$, we have $\frac{v_{0y}^2}{g} = 2H$. Therefore, $R' = R + 2H$.