# Projectile Motion| University Physics

### C) A second rock is thrown straight upward with a speed of 6.000 m/s. If this rock takes 1.636 s to fall to the ground, from what height H was it released?

PART A

In a projectile’s motion, the angle of the initial velocity vi above the horizontal is called the launch angle.

It’s best to place the origin of the coordinate system at ground level below the launching point because in this way all the points of interest (the launching point and the landing point) will have positive coordinates. (Based on your experience, you know that it’s generally easier to work with positive coordinates.) Keep in mind, however, that this is an arbitrary choice. The correct solution of the problem will not depend on the location of the origin of your coordinate system.
Now, define symbols representing initial and final position, velocity, and time. Your target variable is $y_i$, the initial y coordinate of the rock. Your pictorial representation should be complete now, and similar to the picture below:

PART B

ANSWER: $y_i=3.30\:m$

The time Δt needed to move horizontally to the final position $x_f=d=17.0\:m$ is the same time needed for the rock to rise from the initial position $y_i$ to the peak of its trajectory and then fall to the ground. Use the information you have about motion in the horizontal direction to solve for Δt. Knowing this time will allow you to use the equations of motion for the vertical direction to solve for $y_i$.

Find the time spent in the air

$x_f=x_i+v_{ix}\Delta t$

$17=0+\left(12\:cos\left(30^{\circ} \right)\right)\left(\Delta t\right)$

$\Delta \:t=\frac{17}{12\:cos\left(30^{\circ }\:\right)}$

$\Delta \:t=1.64\:s$

Find the initial y position

$y_f=y_i+v_{oy}\Delta t-\frac{1}{2}g\left(\Delta t\right)^2$

$0=y_i+\left(12\:sin\:30^{\circ} \right)\left(1.64\right)-\frac{1}{2}\left(9.800\right)\left(1.64\right)^2$

$y_i=\frac{1}{2}\left(9.800\right)\left(1.64\right)^2-\left(12\:sin\:30^{\circ }\:\right)\left(1.64\right)$

$y_i=3.3\:m$

PART C

ANSWER: $H=3.3\:m$

Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. Because both rocks were thrown with the same initial vertical velocity, $v_{iy}=6.000\:m/s$, and fell the same vertical distance of 3.30 m, they were in the air for the same amount of time. This result was expected and helps to confirm that you did the calculation in Part B correctly.