(a) A snowball rolls off a barn roof that slopes downward at an angle of \(40^\circ\). The edge of the roof is 14.0m above the ground and the snowball has a speed of 7.00m/s as it rolls off the roof. How far from the edge of the barn does the snowball strike the ground if it doesnt strike anything else while falling?

(b) A man 1.9m tall is standing 4.0m from the edge of the barn. Will he be hit by the snowball?


The snowball follows projectile motion.

In part(a), the vertical motion determines the time in air.
The acceleration $$a_x=0$$, $$a_y=+9.80m/s^2$$ and $$v_x=v_0cos\theta_0=7cos40^\circ=5.36m/s$$
Vertical component of velocity $$ v_y=v_0sin\theta_0=4.50m/s$$
Now, distance from ground \(s=14m\).
$$ 14=(4.50)t+(\frac{1}{2}\times9.8\times t^2)$$
We use Sreedhar-Acharya’s method to solve for t.
$$ t=\frac{(-4.5 \pm \sqrt{4.5^2-4(4.9)(-14)}}{2(4.9))}$$
The positive root for \( t=1.29s\)
The horizontal distance$$ v_xt+(1/2)a_x t^2= 6.91m$$

(b) $$ x-x_0=v_xt+\frac{1}{2}a_xt^2$$ gives $$ t=\frac{x-x_0}{v_x}=\frac{4.0}{5.36}=0.746s$$ In this time the snowball travels downward a distance $$ y-y_0=v_y+\frac{1}{2}a_y t^2=6.08m$$ and is therefore
$$ 14.0-6.08=7.9m$$ above the ground. The snowball passes well above the man and does not hit him.