Question

A particle is ejected from the tube at $A$ with a velocity $v$ at an angle $\theta$ with the y-axis as shown in figure. A strong horizontal wind gives the particle a constant horizontal acceleration $a$ in the x-direction. If the particle strikes the ground at a point directly under its released position and the downward $y$ acceleration is taken as $g$, find the height h of the particle.

• A. $h=\frac{2v^2\sin \theta \cos \theta }{a}$
• B. $h=\frac{2v^2\sin \theta \cos \theta }{g}$
• C. $h=\frac{2v^2}{g}\sin \theta (\cos \theta +\frac{a}{g}\sin \theta )$
• D. $h=\frac{2v^2}{a}\sin \theta (\cos \theta +\frac{g}{a}\sin \theta )$

Answer 1

The correct option is D $h=\frac{2v^2}{a}\sin \theta (\cos \theta +\frac{g}{a}\sin \theta )$

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