Question

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.

Answer 1

`y = (2v^2g)/(a^2)`

Step-by-step 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

`\Delta x = v_x t + (1)/(2)a_x t^2`

`0 = v t - (1)/(2)at^2`

`t = (2v)/(a)`

now in the same time it travels in y direction

so the displacement is given as

`y = (1)/(2)gt^2`

`y = (1)/(2)g((2v)/(a))^2`

`y = (2v^2g)/(a^2)`