Projectile Motion

A ball is thrown from the top of a building with height $h_0 = 20$ m at an initial velocity $v_0 = 15$ m/s at an angle $\theta = 30°$ above the horizontal.

Find the horizontal distance the ball travels before hitting the ground. Use $g = 10$ m/s².

Solution

First, find the velocity components:

$v_{0x} = v_0\cos(30°) = 15 \cdot \frac{\sqrt{3}}{2} = 7.5\sqrt{3} \text{ m/s} \approx 13.0 \text{ m/s}$,

$v_{0y} = v_0\sin(30°) = 15 \cdot \frac{1}{2} = 7.5 \text{ m/s}$.

To find the time when the ball hits the ground, use the vertical position equation: $y = h_0 + v_{0y}t - \frac{1}{2}gt^2$

Setting $y = 0$ when the ball hits the ground:

$0 = 20 + 7.5t - 5t^2$

$5t^2 - 7.5t - 20 = 0$

Dividing by $5$: $t^2 - 1.5t - 4 = 0$

Using the quadratic formula:

$t = \frac{1.5 \pm \sqrt{1.5^2 + 4(4)}}{2} = \frac{1.5 \pm \sqrt{2.25 + 16}}{2} = \frac{1.5 \pm \sqrt{18.25}}{2}$

$t = \frac{1.5 \pm 4.27}{2}$

Taking the positive solution:

$t = \frac{1.5 + 4.27}{2} = \frac{5.77}{2} \approx 2.89 \text{ s}$

Now find the horizontal distance traveled:

$x = v_{0x} \cdot t = 7.5\sqrt{3} \cdot 2.89 \approx 13.0 \cdot 2.89 \approx 37.5 \text{ m}$

Therefore, the ball travels approximately 37.5 meters horizontally before hitting the ground.