An important piece of landing equipment must be thrown to a ship, which is moving at 45.0 cm/s , before the ship can dock. This equipment is thrown at 15.0 m/s at $\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$above the horizontal from the top of a tower at the edge of the water, 8.75 m above the ship’s deck (Fig. P3.52). For this equipment to land at the front of the ship, at what distance D from the dock should the ship be when the equipment is thrown? Ignore air resistance.

The velocity contains two components, one is horizontal component and another one is a vertical.

According to the newton’s laws of motion,

$s=ut+\frac{1}{2}a{t}^2$

Where, $s,u,t,$and are displacement, initial velocity, time and acceleration respectively.

Horizontal velocity, $v=45cm/s=0.450m/s$

Vertical velocity, $u=15m/s$

Angle is $60°$.

Vertical distance, $s=8,75m$

Acceleration due to gravity, $g=9.8m/s^2$

For the vertical motion,

$s=u\sin \theta t+\frac{1}{2}a{t}^2$

Substitute known values in the above equation.

$$\begin{gathered} -8.75m=\left(15m/s\times \sin 60°\times t\right)+\frac{1}{2}\times \left(-9.8m/s^2\right)\times t^2 \\ -8.75m=\left(12.99m/s\times t\right)-\left(4.9m/s^2\times t^2\right) \\ \left(4.9\times t^2\right)-\left(12.99\times t\right)-8.75=0 \\ {t}^2-2.65t-1.78=0 \end{gathered}$$

By using quadratic formula, you will get

$$\begin{gathered} t=\frac{-\left(-2.65\right)\pm \sqrt{{\left(2.65\right)}^2-4\left(1\right)\left(-1.78\right)}}{2\left(1\right)} \\ =\frac{2.65\pm \sqrt{7.0225+7.12}}{2} \\ =\frac{2.65\pm \sqrt{14.145}}{2} \\ =\frac{2.65\pm 3.76}{2} \end{gathered}$$

Since time cann’t be negative, hence considering positive values,

$$\begin{gathered} t=\frac{2.65+3.76}{2} \\ =\frac{6.41}{2} \\ =3.20s \end{gathered}$$

This much time will be needed by equipment to cover that much height.

Now relative velocity of equipment with respect to ship is,

$v_r=u_x-\left(-v_s\right)$

Here, $v_s$is the velocity of the ship in the opposite direction of that of velocity of equipment. Therefore,

$$\begin{gathered} v_r=7.5-\left(-0.45\right) \\ =7.5+0.45 \\ =7.95m/s \end{gathered}$$

Since no acceleration is acting in horizontal direction, the distance will be,

$$\begin{gathered} D=v_{r}t \\ =7.95\times 3.20 \\ =25.44m \end{gathered}$$

Hence, the ship must be 25.44 m away from the dock to land in front of the ship.