Q52P - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

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

Step 1: Introduction:

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.

Step 2: As given data:

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

 

Vertical velocity,  $u=15 m/s$

 

Angle is $60°$ .

 

Vertical distance,  $s=8,75 m$

 

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

Step 3: The distance D from the deck:

For the vertical motion, 

                                                                                    

 $s=u\sin \theta t+\frac{1}{2}a{t}^2$

 

Substitute known values in the above equation.

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

 

 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,

 

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

 

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,

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

 

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

 

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

 

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