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A student sits atop a platform a distance h above the ground. He throws a large firecracker horizontally with a speed v. However, a wind blowing parallel to the ground gives the firecracker a constant horizontal acceleration with magnitude a. As a result, the firecracker reaches the ground directly below the student. Determine the height in terms of v,a, and g. Ignore the effect of air resistance on the vertical motion.

Respuesta :

Answer:

[tex]y = \frac{2v^2g}{a^2}[/tex]

Explanation:

Since fire cracker land at the position just below the student

so we can say that the displacement in X direction must be zero

so we will have

[tex]\Delta x = v_x t + \frac{1}{2}a_x t^2[/tex]

[tex]0 = v t - \frac{1}{2}at^2[/tex]

[tex]t = \frac{2v}{a}[/tex]

now in the same time it travels in y direction

so the displacement is given as

[tex]y = \frac{1}{2}gt^2[/tex]

[tex]y = \frac{1}{2}g(\frac{2v}{a})^2[/tex]

[tex]y = \frac{2v^2g}{a^2}[/tex]

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