Q67P - University Physics with Modern Physics

Question

A cart carrying a vertical missile launcher moves horizontally at a constant velocity of 30.0m/s to the right. It launches a rocket vertically upward. The missile has an initial vertical velocity of 40.0m/s relative to the cart. (a) How high does the rocket go? (b) How far does the cart travel while the rocket is in the air? (c) Where does the rocket land relative to the cart?

Step-by-Step Solution

Answer

(a)  81.6m      

 

(b) 245m

 

(c) Rocket will land on the cart.

Step 1: Newton’s law

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

 

According to newton’s laws of motion,

 

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

 

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

 

For the vertical and horizontal motion, the velocity component will be respectively.

Step 2: Given

Horizontal velocity = 30m/s

 

Vertical velocity = 40m/s

Step 3: (a) Height of rocket

From newton’s laws of motion, (final velocity v is zero)

 

 $$\begin{gathered} v^2=u^2+2as \\ {0}^2={\left(40.0m/s\right)}^2+2\left(9.8m/s\right)s \\ s=81.6m \end{gathered}$$

 

Vertical distance travel by rocket is 81.6m.

Step 4: (b) Distance travel by cart while the rocket is in the air

From vertical motion, when missile return to initial level, the displacement will be zero.

 

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ 0=40m/s\times t\times \frac{1}{2}\times 9.8m/s^2\times t^2 \\ t=8.16s \end{gathered}$$

 

From the horizontal motion,

 

$$\begin{gathered} s=ut \\ s=30.0m/s\times 8.16s \\ s=245m \end{gathered}$$

 

So the distance will be 245m.

Step 5: (c) Landing of rocket relative to the cart

The rocket also traveled 245m horizontally so rocket land into the cart.