### Home > CALC > Chapter 12 > Lesson 12.1.3 > Problem12-39

12-39.

A ball is thrown from a window $25$ meters above the ground with an initial velocity of $40$ meters per second and an angle of inclination of $\frac { \pi } { 6 }$. Let the origin be the point on the (level) ground below the window.

1. Assume the acceleration due to gravity is $−10 \frac{\text{meters}}{\text{sec}^2}$. Find $x(t)$ and $y(t)$.

$\vec{v}(t)=\langle v_0\cos(\theta),v_0\sin(\theta)+a(t)\rangle=\langle 40\cos\Big(\frac{\pi}{6}\Big),40\sin\Big(\frac{\pi}{6}\Big)-10t\rangle$
$x(t)=40t\cos\Big(\frac{\pi}{6}\Big)+C_1\text{ }y(t)=40t\sin\Big(\frac{\pi}{6}\Big)-5t^2+C_2$

2. Find the angle at which the ball hits the ground.

First, determine when the ball hits the ground. This is when $y(t) = 0$.
Then angle is the angle created by the slope.
Think of drawing a slope triangle and calculating the angle.