Question

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

• A. $\left(\text{R} + \text{H}\right) \text{, } \frac{\text{H}}{2}$
• B. $\left(\text{R} + \frac{\text{H}}{2}\right) \text{, } 2 \text{H}$
• C. (R + 2H), H
• D. (R + H), H

Answer 1

Answer: (D)

$\text{T} = \frac{2 \text{u}_{\text{y}}}{\text{g}} \text{, } \text{H} = \frac{\text{u}_{\text{y}}^{2}}{2 \text{g}}$
where T = time of flight of the projectile and u_x and u_y are respectively the horizontal and vertical components of initial velocity
When a horizontal acceleration is also given to the projectile u_y and hence T and H will remain unchanged while the range will become
$\text{R} = \text{u}_{\text{x}} \text{T} + \frac{1}{2} \text{aT}^{2}$
$=\mathrm{R}+\frac{1}{2}\left(\frac{\mathrm{g}}{4}\right)\left(\frac{4 \mathrm{u}_{\mathrm{y}}^2}{\mathrm{~g}^2}\right)$
= R + H