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 $θ$ 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.

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Solution

Let the height of the walls be $h$
$h = (u s i n θ) t - 0.5 g t^{2}$
or
$g t^{2} - 2 (u s i n θ) t + 2 h = 0... (1)$
The projectile attains a ht $h$ at times $t_{1} t_{2}$

From the above quadratic equation sum of roots
$t_{1} + t_{2} = \frac{2 u s i n θ}{g} t_{1} t_{2} = \frac{2 h}{g}$
a.
Thus the height of wall $h = \frac{g t_{1} t_{2}}{2}$
b.
Solving the quadratic equation,
$t_{1} = \frac{u s i n θ - \sqrt{u^{2} {s i n}^{2} θ - 2 g h}}{g}$
$t_{2} = \frac{u s i n θ + \sqrt{u^{2} {s i n}^{2} θ - 2 g h}}{g}$
Range,
$R = \frac{u^{2} s i n 2 θ}{g}$
and
The distance between walls.
$D = u c o s θ [t_{2} - t_{1}] = u c o s θ \times \frac{2 \sqrt{u^{2} {s i n}^{2} θ - 2 g h}}{g}$

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