Question

A ball is thrown eastward across level ground. A wind blows horizontally to the east, and assume that the effect of wind is to provide a constant force towards the east, equal in magnitude to the weight of the ball. The angle $\theta$ so that it travels maximum horizontal distance is

• A. ${45}^{\circ }$
• B. ${37}^{\circ }$
• C. ${53}^{\circ }$
• D. ${67.5}^{\circ }$

Answer 1

Answer: (D)

${67.5}^{\circ }$

The acceleration is g both horizontally and vertically.
time of flight $=t=\frac{2usin}{g}$
horizontally s $=ut+\frac{1}{2}a{t}^2$
$s=\frac{ucos\theta \left(2usin\theta \right)}{g}+\left(0.5\right)g(\frac{2usin\theta }{g}{)}^2$
$s=\left(\frac{1}{g}\right)[u^{2}sin2\theta +2u^{2}si{n}^2\theta ]$
for maximum ; s $=\frac{ds}{d\theta }=0$
$\frac{u^2}{g}[2cos2\theta +4sin\theta cos\theta ]=0$
$[cos2\theta +sin2\theta ]=0ortan2\theta =-1$
$\theta =\frac{{135}^o}{2}={67.5}^o$
so,angle the ball should be thrown so that it travels the
maximum horizontal distance $={67.5}^o$
hence, option $D$ is correct.