A boy throws a ball upward with a speed of 12 m/s. The wind imparts a horizontal acceleration of 0.4 m / s 2 . At what angle θ to the vertical, the ball must be thrown so that it returns to the point

Question

A boy throws a ball upward with a speed of 12 m/s. The wind imparts a horizontal acceleration of 0.4 m / s 2 . At what angle θ to the vertical, the ball must be thrown so that it returns to the point of release.

TestHub · Accepted Answer

Return to the release point needs zero net horizontal displacement:
$v_{x} t = \frac{1}{2} (0.4) t^{2}$ ,
so
$v_{x} = 0.2 t$ ;
vertically
$t = \frac{2 v _{y}}{g}$ .
Hence
$tan θ = \frac{v _{x}}{v _{y}} = \frac{0.2 \times 2}{g} = \frac{0.4}{g} = \frac{1}{25}$
for
$g = 10 m / s^{2}$ .
Matches the printed answer. (Check:
$v_{x}^{2} + v_{y}^{2} = (0.2 t)^{2} + (5 t)^{2} = 144$
gives
$t = 2.4 s, v_{x} = 0.48 m / s, v_{y} = 11.99 m / s$
- consistent with the 12 m/s throw.)

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