Q70P - University Physics with Modern Physics

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

Step-by-Step Solution

Answer

$h=\frac{2v^{2}g}{a^2}$ is the height h in terms of v, a, and g by Ignoring the effect of air resistance on the vertical motion.

Step 1: Newton’s law

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.

Step 2: Given

Horizontal speed = v

 

Horizontal acceleration = a

 

Vertical acceleration = g

 

Height = h

Step 3: 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.