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(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?

Solution:

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.