A projectile is fired with an initial speed of$36.6\mathrm{m}/\mathrm{s}$at an angle of$42.2^{\mathrm{o}}$above the horizontal on a long, flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and (d) the speed of the projectile 1.50 s after firing.

The motion of an object thrown in the air at some angle is known as the projectile motion. It takes place under the effect of the earth’s acceleration due to gravity.

The horizontal and vertical motions of a projectile can be determined independently.

Let the positive y-direction be upwards, and the origin be the point of projectile launch.

The magnitude of the initial velocity of the projectile is $u=36.6\mathrm{m}/\mathrm{s}$.

The launching angle of the projectile is $\theta =42.2^{\mathrm{o}}$.

The acceleration experienced by the projectile along the vertical is $a_{\mathrm{y}}=-g=-9.8\mathrm{m}/{\mathrm{s}}^2$.

The vertical velocity of the projectile at the top is 0.

The vertical component of the initial velocity is

$\begin{array}{c}{u}_{\mathrm{y}}=u\sin \theta \\ =\left(36.6\mathrm{m}/\mathrm{s}\right)\left(\sin {42.2}^{\mathrm{o}}\right)\end{array}$

The maximum height is

$\begin{array}{c}H=\frac{u^2{\sin }^2\theta }{2g}\\ =\frac{{\left(36.6\mathrm{m}/\mathrm{s}\right)}^2\left({\sin }^2{42.2}^{\mathrm{o}}\right)}{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\\ =30.84\mathrm{m}\end{array}$

Thus, the maximum height achieved by the projectile is 30.84 m.

The total vertical displacement for the entire flight is zero as the projectile reaches the ground. The total time of flight is given by

$\begin{array}{c}T=\frac{2u\sin \theta }{g}\\ =\frac{2\left(36.6\mathrm{m}/\mathrm{s}\right)\left(\sin 42.2^{\mathrm{o}}\right)}{9.8\mathrm{m}/{\mathrm{s}}^2}\\ =5.02\mathrm{s}\end{array}$

On the other hand, $t=0\mathrm{s}$corresponds to the time when the displacement is zero.

The range of the projectile is the total horizontal distance covered by it.The component of velocity along the positive x-axis is given by

$\begin{array}{c}{u}_{\mathrm{x}}=u\cos \theta \\ =\left(36.6\mathrm{m}/\mathrm{s}\right)\left(\cos 42.2^{\mathrm{o}}\right)\end{array}$

The total horizontal distance covered by the projectile is the horizontal velocity multiplied by the time.

$\begin{array}{c}x=u_{\mathrm{x}}t\\ =\left(36.6\mathrm{m}/\mathrm{s}\right)\left(\cos 42.2^{\mathrm{o}}\right)\left(5.02\mathrm{s}\right)\\ =136.11\mathrm{m}\end{array}$

Thus, the range of the projectile is 136.11 m.

There are two components of the projectile’s velocity 1.50 s after firing. The horizontal velocity remains constant throughout and is given by

$\begin{array}{c}{u}_{\mathrm{x}}=u\cos \theta \\ =\left(36.6\mathrm{m}/\mathrm{s}\right)\left(\cos 42.2^{\mathrm{o}}\right)\\ =27.11\mathrm{m}/\mathrm{s}\end{array}$

The vertical velocity is determined by

$\begin{array}{c}{u}_{\mathrm{y}}=u\sin \theta +at\\ =u\sin \theta -gt\end{array}$

Substituting the numerical values,

$\begin{array}{c}{u}_{\mathrm{y}}=\left(36.6\mathrm{m}/\mathrm{s}\right)\sin 42.2^{\mathrm{o}}-\left(9.8\mathrm{m}/\mathrm{s}\right)\left(1.5\mathrm{s}\right)\\ =9.89\mathrm{m}/\mathrm{s}\end{array}$

Therefore, the speed of the projectile is given by the vector sum of the horizontal and vertical velocities.

$\begin{array}{c}u\text{'}=\sqrt{u_{\mathrm{x}}^2+u_{\mathrm{y}}^2}\\ =\sqrt{{\left(27.11\mathrm{m}/\mathrm{s}\right)}^2+{\left(9.89\mathrm{m}/\mathrm{s}\right)}^2}\\ =28.9\mathrm{m}/\mathrm{s}\end{array}$

Thus, the speed of the projectile 1.50 s after firing is $28.9\mathrm{m}/\mathrm{s}$.