91Ó°ÊÓ

A projectile motion is a curved path traveled by an object launched in the air, moving under constant acceleration.For example, archery, a javelin throw, water escaping a hose, etc.

The total time taken by the dart to reach the ground is $T=4\mathrm{s}$.

During the vertical motion of the dart, half the time of flight $\left(t_{\mathrm{up}}\right)$is spent moving upward. The time taken by the dart to fall back on the ground is $t_{\mathrm{down}}$.

Thus, $t_{\mathrm{up}}=t_{\mathrm{down}}=\frac{T}{2}$.

The time taken by the upward flight is $t_{\mathrm{up}}=2\mathrm{s}$.

Let the upward direction be the positive y-direction at the top of the flight.

The final vertical velocity of the dart at the top of its flight is $v_{\mathrm{y}}=0$.

The acceleration experienced by the dart while going upward is $a=-g=-9.8\mathrm{m}/{\mathrm{s}}^2$.

The initial or the launching velocity is the muzzle speed at which the gun fires the dart. It is given by:

$\begin{array}{c}{v}_{\mathrm{y}}=u_{\mathrm{y}}+a{t}_{\mathrm{up}}\\ 0=u_{\mathrm{y}}-g{t}_{\mathrm{up}}\\ u_{\mathrm{y}}=-\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2\mathrm{s}\right)\\ =19.6\mathrm{m}/\mathrm{s}\end{array}$

Thus, the initial velocity of the dart is $19.6\mathrm{m}/\mathrm{s}$.

The maximum horizontal range of the gun is determined at an angle of $\mathrm{θ}={45}^{\mathrm{o}}$. The formula for the maximum range is given by:

$\begin{array}{c}R=\frac{u_{\mathrm{y}}^2\sin 2\mathrm{θ}}{g}\\ =\frac{{\left(19.6\mathrm{m}/\mathrm{s}\right)}^2\sin 2\left({45}^{\mathrm{o}}\right)}{9.8\mathrm{m}/{\mathrm{s}}^2}\\ =39.2\mathrm{m}\end{array}$

Thus, the gun’s maximum horizontal range is 39.2 m.