Projectile Motion| University Physics

A rock is thrown with a speed of 12.0 m/s and a launch angle of 30.0° (above the horizontal) travels a horizontal distance of d=17.0\:m before hitting the ground. Use the value g=9.800\:m/s^2 for the free-fall acceleration.

A) Which diagram represents an accurate sketch of the rock’s trajectory?

B) Find the height y_i from which the rock was launched.

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?




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:

The figure shows a trajectory of a rock being thrown at an angle theta above the horizontal from the initial point above the ground. Acceleration points vertically downward. The rock travels a horizontal distance d before it hits the ground at the final point. The trajectory is a concave downward curve. The rock’s velocity at the moment when it hits the ground is tangent to the trajectory. The initial and final points are characterized by time, horizontal and vertical positions, and corresponding components of velocity.


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)



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.