91Ó°ÊÓ

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.

The slope of the barn roof is$\mathrm{θ}=45°$ .

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 .

(a)

Calculate the y component of the velocity.

$v_{0y}=v_0\sin \mathrm{θ}$

Substitute 7 m/s for$v_0$ and$40°$ for$\mathrm{θ}$ in the above equation.

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

Calculate the x component of the velocity.

$v_{0y}=v_0\sin \mathrm{θ}$

Substitute 7 m/s for$v_0$ and$40°$ for$\mathrm{θ}$ 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,

$t=1.28s$

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.

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

Therefore 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 below,

(c)

Now, for the horizontal motion, gravity is zero

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

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

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

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.