Q59P - University Physics with Modern Physics

Question

A snowball rolls off a barn roof that slopes downward at an angle of $\mathbf{40}\mathbf{°}$ (Fig. P3.59) The edge of the roof is 14.0 m above the ground, and the snowball has a speed of 7.0m/s as it rolls off the roof. Ignore air resistance. (a) How far from the edge of the barn does the snowball strike the ground if it doesn’t strike anything else while falling? (b) Draw $\mathit{x}\mathbf{-}\mathit{t}\mathbf{,}\mathit{y}\mathbf{-}\mathit{t}\mathbf{,}{\mathit{v}}_{\mathbf{x}}\mathbf{-}\mathit{t}\mathbf{,}{\mathit{v}}_{\mathbf{y}}\mathbf{-}\mathit{t}$ graphs for the motion in part (a). (c) A man 1.9 m tall is standing from the edge of the barn. Will the snowball hit him?

                                                            

Step-by-Step Solution

Answer

a) The edge of the barn will strike the ground at 6.86 m and the snowfall will not hit him.

b) The graph is shown in step 4.

c) The snow ball will not hit the man.

Step 1: Introduction

The second law of motion is given by,

 

 $∆y=v_{0y}t+\frac{1}{2}g{t}^2$...... (1)

 

Here $∆y$ is the distance between edge of the roof and the ground, $v_{0y}$ is the velocity along the y direction, t is the time and g is the acceleration due to gravity.

Step 2: Given data

The slope of the barn roof is $\theta =45°$ . 

 

Edge of the roof is 14.0 m . 

 

The initial speed of the snowball is u = 7 m/s . 

 

The height of the man is 1.9 m . 

 

The distance of the man from the edge of the barn is 4.0 m . 

Step 3: Distance of the edge of the barn

(a)

 

Calculate the y component of the velocity.

 

$v_{0y}=v_0\sin \theta$ 

Substitute 7 m/s for $v_0$ and $40°$ for $\theta$ in the above equation.

$$\begin{gathered} v_{0y}=7\sin 40° \\ =4.5m/s \end{gathered}$$

 

Calculate the x component of the velocity.

$v_{0y}=v_0\sin \theta$ 

Substitute 7 m/s for $v_0$ and $40°$ for $\theta$ in the above equation.

$$\begin{gathered} v_{0y}=7 \cos 40° \\ =5.36m/s \end{gathered}$$

Using the equation (1)

$∆y=v_{0y}t+\frac{1}{2}g{t}^2$

Substitute 14.0 m for $∆y$ , 4.5 m/s for $v_{0y}$ and 9.8 m/$s^2$ for g in the above equation.

 $14=4.5t+\frac{1}{2}g{t}^2$

On solving the above equation we get, 

 $t=1.28 s$

 

Now use the equation (1) in terms of x ,

$x-x_0=v_{0x}t+\frac{1}{2}{a}_x{t}^2$

Substitute 5.36m/s for $v_{0x}$ , 1.28 s for t and $0m/s^2$ for $a_x$ in the above equation.

$$\begin{gathered} x-x_0=\left(5.36\right)\left(1.28\right)+\frac{1}{2}\left(0\right){\left(1.28\right)}^2 \\ =6.86 m \end{gathered}$$

Therefore the edge of the barn will strike the ground at 6.86 m and the snowfall will not hit him.

Step 4: Graph of x - t , y - t , v x - t , v y - t

(b)

 

The graph is shown below,

Step 5: Will the snowball hit the man?

(c)

 

Now, for the horizontal motion, gravity is zero

                                                                                           

$$\begin{gathered} s=u\cos \theta t+\frac{1}{2}a{t}^2 \\ 4m=5.36m/s\times t+\frac{1}{2}\times 0m/s^2\times {\left(t\right)}^2 \\ t=0.746s \end{gathered}$$

 

In this time the snowfall travels downward a distance that is by vertical motion,

 

$$\begin{gathered} s=u \sin \theta t+\frac{1}{2}a{t}^2 \\ s=4.5m/s\times 0.746s+\frac{1}{2}\times 9.8m/s^2\times {\left(0.746s\right)}^2 \\ s=7.012m \end{gathered}$$

 

Therefore, the difference in distance is,

 

$$\begin{gathered} d=14.0m-6.08m \\ =6.988m \\ =7m \end{gathered}$$

 

Therefore the snowball passes above the man but it will not hit him.