## Projectile Motion Problems

Two projectile motion problems are solved, demonstrating the use of the equations for motion at constant acceleration, and breaking the velocity vector up into its x and y components.

Equations of Motion at Constant Acceleration:
$v = v_o + at$

$x = x_o + v_o t + \frac{1}{2} a t^2$

$v^2 = v_o^2 + 2ad$

 A ball is shot horizontally at 20 m/s, off the edge of a cliff. Determine the distance from the top of the cliff, and the velocity, after 2 seconds.

Distance
x-direction
The velocity is constant in the x-direction, so the distance is:

$d = vt = (20 m/s) (2 sec) = 40 m$

y-direction
The ball is accelerating due to gravity in the y-direction
The equation of constant acceleration that determines the distance is:
$y = y_o + v_1 t + \frac{1}{2} a t^2$

With the initial velocity in the y-direction = 0;
The initial height = 0.

$y = 0 + 0 + \frac{1}{2} (-9.8 m/s^2) (2 sec)^2$
$y = -19.6 m$

With the x and y components of the distance from the cliff, the distance is determined:
$d = \sqrt{x^2 + y^2} = \sqrt{(40m)^2 + (-19.6m)^2} = 44.5 m$

Velocity
x-direction
The velocity in the x-direction remains constant.
$v_x = 20 m/s$

y-direction
The initial velocity in the y-direction is $v_{y1} = 0$ because the initial velocity was totally in the x-direction. In the y-direction, it's just like the ball was dropped off the cliff.
The equation of constant acceleration that will determine the final velocity in the y-direction, after 2 seconds, is:

$v_{y2} = v_{y1} + at$

$v_{y2} = 0 + (-9.8m/s) (2 sec) = -19.6 m/s$

With the x and y components of the velocity, the final velocity can be calculated:

$v = \sqrt{v_x^2 + V_y^2} = \sqrt{(20 m/s)^2 + (-19.6 m/s)^2} = 28.0 m/s$

The angle of the velocity:
$tan \theta = \frac{v_y}{v_x} = \frac{-19.6 m/s}{20 m/s}$

$\theta = -44.4^o$

So, the angle is $44.4^o$ down from horizontal.

Projectile Motion Problems - #2

 A ball is shot off a 40 m high cliff at 42 m/s at an angle of 30 degrees.
• Find the maximum height above the cliff
• Find the magnitude of the final velocity just before it hits the ground
• Find the distance from the cliff to where it lands

Here's a reminder of the main
Equations of Motion at Constant Acceleration:
$v = v_o + at$

$x = x_o + v_o t + \frac{1}{2} a t^2$

$v^2 = v_o^2 + 2ad$

Maximum height above the cliff
Break the initial velocity into its x and y components:
$v_x = v_o \cos \theta$
$v_y = v_o \sin \theta$

We use only the y-component of the velocity to find the maximum height. As the ball goes up, the velocity in the y-direction slows down due to gravity, until at the peak of its trajectory, the y-component of the velocity is 0.
Using the first to the three equations for constant acceleration, we can find the time for the ball to reach the maximum height, given that $v_{y2} = 0$.

$v_2 = v_1 + at$

$t = \frac{(v_2 - v_1)}{a} = \frac{0 - v_o \sin \theta}{a}$

"a" is the acceleration due to gravity: $a = -9.8 m/s^2$
So, the time to reach the maximum height is:

$t = \frac{0 - v_o \sin \theta}{a} = \frac{-(42 m/s) \sin 30^o}{-9.8 m/s^2} = 2.1 sec$

With this time, we can use the second equation for constant acceleration, listed above, to determine the maximum height.
$y = y_o + v_o t + \frac{1}{2} a t^2$

With the top of the cliff at $y_o = 0$, the height above the cliff is:
$y = 0 + (42 m/s) \sin 30^o (2.1 s) + \frac{1}{2} (-9.8 m/s^2) (2.1 s)^2$

$y = 22.5 m$

Magnitude of final velocity
x-component of the final velocity
The velocity in the x-direction is constant:
$v_x = v_o \cos \theta = (42 m/s) \cos 30^o = 36.4 m/s$

y-component of the final velocity
We know the height of the cliff and how high it goes to maximum height where the velocity in the y-direction is 0. From that height, it's just free-fall. We can use the 3rd equation for constant acceleration to determine the final velocity in the y-direction.

Total distance, d:
d = cliff height + max. height = $40 m + 22.5 m = 62.5 m$

3rd equation for constant acceleration:
$v^2 = v_o^2 + 2ad$

$v_{y2}^2 = 0 + 2 (-9.8 m/s^2)(62.5 m)$

$v_{y2} = -35 m/s$
(The negative sign just determines the direction of the vector, so does not conflict with taking the square root.)

The magnitude of the velocity is:
$v = \sqrt{v_x^2 + v_y^2} = \sqrt{ (36.4 m/s)^2 +(-35 m/s)^2} = 50.5 m/s$

Distance from the cliff to where it lands
Now we need to calculate the time it takes for the ball to drop from it's maximum height, so we will have the total time that the ball is in the air. With the final velocity in the y-direction, from the previous question, we can use the 1st equation for constant acceleration to determine the time:
$v = v_o + at$
$t = \frac{v_{y2} - v_{y1}}{a} = \frac{-35 m/s - 0 }{-9.8 m/s^2} = 3.6 s$

So the total time that the ball is in the air is:
t = time from cliff to max. height + time from max. height to hitting the ground
$t = 2.1 s + 3.6 s = 5.7 s$

Since the velocity in the x-direction is constant, we can use that velocity with the total time, to determine the distance:
$d = v_x t = (36.4 m/s) (5.7 s) = 207.5 m$