Question

A particle is projected from the ground at $t=0$ so that on its way it just clears two vertical walls of equal height on the ground. The particle was projected with initial velocity $u$ and at angle $\theta$ with the horizontal. If the particle passes just grazing top of the wall at time $t=t_1$ and $t=t_2$ then calculate.
a. the height of the wall.
b. the time $t_1$ and $t_2$ in terms of height of the wall. Write the expression for calculating the range of this projecting and separation between the walls.

Answer 1

Let the height of the walls be $h$

$h=\left(usin\theta \right)t-0.5g{t}^2$

or

$g{t}^2-2\left(usin\theta \right)t+2h=\mathrm{0...}\left(1\right)$

The projectile attains a ht $h$ at times $t_1{t}_2$

From the above quadratic equation sum of roots

$t_1+t_2=\frac{2usin\theta }{g}{t}_1{t}_2=\frac{2h}{g}$

a.

Thus the height of wall $h=\frac{g{t}_1{t}_2}{2}$

b.

Solving the quadratic equation,

$t_1=\frac{usin\theta -\sqrt{u^2{sin}^2\theta -2gh}}{g}$

$t_2=\frac{usin\theta +\sqrt{u^2{sin}^2\theta -2gh}}{g}$

Range,

$R=\frac{u^{2}sin2\theta }{g}$

and

The distance between walls.

$D=ucos\theta [t_2-t_1]=ucos\theta \times \frac{2\sqrt{u^2{sin}^2\theta -2gh}}{g}$