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 h in terms of v, a, and g. Ignore the effect of air resistance on the vertical motion.

The velocity contains two components, one is horizontal component and another one is vertical.

According to the newton’s laws of motion,

$s=ut+\frac{1}{2}g{t}^2$

Where, s, u, t, g, are displacement, initial velocity, time and acceleration.

For the vertical and horizontal motion the velocity component will be$u\sin \theta \text{and cos}\theta$
respectively.

Horizontal speed = v

Horizontal acceleration = a

Vertical acceleration = g

Height = h

The firecracker’s falling vertical motion, here initial velocity is zero

$$\begin{gathered} s=ut+\frac{1}{2}g{t}^2 \\ h=0\times t+\frac{1}{2}g{t}^2 \\ t=\sqrt{\frac{2h}{g}} \end{gathered}$$

The firecracker’s horizontal motion at the time t, here taking student position or distance zero, a is acceleration.

$$\begin{gathered} s=ut+\frac{1}{2}g{t}^2 \\ 0=v\times t-\frac{1}{2}a{t}^2 \\ \frac{2v}{a}=\sqrt{\frac{2h}{g}} \\ h=\frac{2v^{2}g}{a^2} \end{gathered}$$

This is the height h in terms of v, a, and g by Ignoring the effect of air resistance on the vertical motion.