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Let's discuss the beautiful snowball problem useful for Physics Olympiad.

**The Snowball Problem:**

(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 doesn't 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 the projectile motion.

In part(a), the vertical motion determines the time in the 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.

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