91Ó°ÊÓ

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$u\sin \mathrm{θ}\text{and cos}\mathrm{θ}$
respectively.

Horizontal speed = v

Horizontal acceleration = a

Vertical acceleration = g

Height = h

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

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

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

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

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