### Home > APCALC > Chapter 12 > Lesson 12.1.3 > Problem12-36

12-36.

A ball is thrown from a window $25$ meters above the ground with an initial velocity of $40$ m/sec 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 $–10$ m/sec2. Write equations for $x(t)$ and $y\left(t\right)$.

$\vec{v}(t)=\langle v_0\cos(\theta),v_0\sin(\theta)+a(t)\rangle=\langle 40\cos\Big(\frac{\pi}{6}\Big),40\sin\Big(\frac{\pi}{6}\Big)-10t\rangle$

$x(t)=40t\cos\Big(\frac{\pi}{6}\Big)+C_1\text{ }y(t)=40t\sin\Big(\frac{\pi}{6}\Big)-5t^2+C_2$

$x(0)=0\text{ }y(0)=25$

2. Determine 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.

$\text{slope}=\frac{y^\prime(t)}{x^\prime(t)}$