Q50P - University Physics with Modern Physics

Question

A cannon, located 60.0 m from the base of a vertical 25.0 m tall cliff, shoots a 15 kg shell at $\mathbf{43}\mathbf{.}\mathbf{0}\mathbf{°}$ above the horizontal toward the cliff. (a) What must the minimum muzzle velocity be for the shell to clear the top of the cliff? (b) The ground at the top of the cliff is level, with a constant elevation of 25.0 m above the cannon. Under the conditions of part (a), how far does the shell land past the edge of the cliff?

Step-by-Step Solution

Answer

1. The minimum muzzle velocity and time will be $u=32.6 m/s$ . 
2. The shell will land at a distance of 60 m . 

Step 1: Introduction

If a particle is thrown in projectile motion, then the velocity of the particle will contain two components, first one is horizontal and other one is vertical.

 

According to the Newton’s laws of motion,

 

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

 

Here s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 

From the newton’s law of motion

 

$v=u+at$

 

Here, $v,u,a,t$ are final velocity, initial velocity, acceleration due to gravity, and time.

Step 2: Given data

Vertical distance =  25 m   

 

Horizontal distance =   60 m           

 

Weight of the shell =  15 kg

                                                                                           

Angle = $43°$

Step 3: Minimum muzzle velocity

(a)

 

For horizontal motion gravity is zero and velocity is $u\cos \theta$ and for vertical motion velocity is $u\sin \theta$ .

 

For the horizontal motion from the law of motion of Newton,  

 

$$\begin{gathered} s=u\cos \theta t+\frac{1}{2}a{t}^2 \\ 60=u\cos 43°\times t+\frac{1}{2}\times 0\times {\left(t\right)}^2 \\ t=\frac{60 }{u \cos 43°} \end{gathered}$$

 

In the time t the vertical displacement is given by,

$$\begin{gathered} s=u\sin \theta t+\frac{1}{2}a{t}^2 \\ 25=u\sin 43°\times t+\frac{1}{2}\times -9.8\times t^2 \end{gathered}$$

Substitute $\frac{60m}{u\cos 43°}$  for t in the above equation and solve for u .

 

On solving the above equation we get,

 

$$\begin{gathered} 25=u\sin 43°\times \left(\frac{60}{u\cos 43°}\right)+\frac{1}{2}\times -9.8\times {\left(\frac{60}{u\cos 43°}\right)}^2 \\ u=32.6 m/s \end{gathered}$$

 

Now, substitute 32.6 m/s for u in the equation of time t .

 

$$\begin{gathered} t=\frac{60}{32.6\cos 43°} \\ =2.516s \\ \approx 2.52s \end{gathered}$$

 

Therefore the minimum muzzle velocity for the shell to clear the top of the cliff is 32.6 m/s . 

Step 4: The distance by the shell land past the edge of the cliff

(b)

 

The equation for the horizontal displacement of the canon is,

 

 $x=\left(u\cos \theta \right)t$

 

Substitute for , for and for in the above equation.

 

$$\begin{gathered} x=\left(32.6\cos 43°\right)\left(2.52\right) \\ =60 m \end{gathered}$$

 

Therefore the shell will land at a distance of 60 m .