Question

A ball is released from a window that is 10 meters above the ground. When
the ball leaves your hand, it is moving 8m/s at an angle 30 degrees below the horizontal.
Ignoring the effect of air resistance, how far horizontally from the window will the
ball hit the ground?

Answer 1

Let's first find $x$ and $y$ components of the initial velocity of the ball:

$$v_{0x}=v_0cos\theta=8\ \dfrac{m}{s}\cdot cos30^{\circ}=6.93\ \dfrac{m}{s},$$ $$v_{0y}=v_0sin\theta=8\ \dfrac{m}{s}\cdot sin30^{\circ}=4\ \dfrac{m}{s}.$$

Then, we can find the time that the ball takes to reach the ground from the kinematic equation. Let's take the upwards as the positive direction, then we get:

$$y=y_0+v_{0y}t-\dfrac{1}{2}gt^2,$$ $$0=10-4t-4.9t^2,$$ $$4.9t^2+4t-10=0.$$

This quadratic equation has two roots: $t_1=1.08$ and $t_2=-1.89$. Since time can't be negative the correct answer is $t=1.08\ s$.

Finally, we can find the range of the ball:

$$x=v_{0x}t=6.93\ \dfrac{m}{s}\cdot1.08\ s=7.48\ m.$$