Question

A student sits atop a platform a distance h above the ground. He throws a large firecracker horizontally with a speed. 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.

Answer 1

Answer:

The height h, in terms of v, a, and g is h = 2·g·$(\frac{v}{a} )^2$

Explanation:

To find the solution, we list out the variables and their relationships

Height from which fire cracker is thrown = h

Speed with which fire cracker is thrown = v

Acceleration of wind in opposite direction of v = a

We have horizontal motion of the firecracker given by

S = v·t + 0.5·a·t² since a is in opposite direction to v, we have

S = v·t - 0.5·a·t² also since the firecracker comes to rest at the ground directly below the student we have displacement S = 0

Therefore S = v·t - 0.5·a·t² ⇒ 0 = v·t - 0.5·a·t², so that  0.5·a·t² = v·t or t = $\frac{v}{0.5*a}$

= $\frac{2v}{a}$

For the vertical motion, we have by Newton's law of motion

h = u·t + $\frac{1}{2}$·g·t² however, u = 0

∴ h =  $\frac{1}{2}$·g·t² and substituting the value for t from the horizontal motion calculation solution, we have

h =  $\frac{1}{2}$·g·$(\frac{2v}{a})^2$ = where g = 9.81 m/s², we have h = 19.62 $(\frac{v}{a} )^2$ in terms of g we have h = 2·g·$(\frac{v}{a} )^2$