Projectile Motion Example
Probably the simplest way to see how to apply these constant acceleration equations is to work through a sample problem. Let's say you're on top of a cliff, which drops vertically 150 m to a flat valley below. You throw a ball off the cliff, launching it at 8.40 m/s at an angle of 20¡ above the horizontal.
(a) How long does it take to reach the ground?
(b) How far is it from the base of the cliff to the point of impact?
It's a good idea to be as systematic as possible when it comes to analyzing the situation. Here's an example of how to organize the information you're given. First, draw a diagram.
~
Then set up a table to keep track of everything you know. It's important to pick an origin, and to choose a coordinate system showing positive directions. In this case, the origin was chosen to be the base of the cliff, with +x being right and +y being up. You don't have to choose the origin or the positive directions this way. Pick others if you'd like, and stick with them (an origin at the top of the cliff, and/or positive y-direction down would be two possible changes).
~
Now that everything's neatly organized, think about what can be used to calculate what. You know plenty of y-information, so we can use that to find the time it takes to reach the ground. One way to do this (definitely not the only way) is to do it in two steps, first calculating the final velocity using the equation:
~
This gives vy2 = 2.8732 + 2 (-9.8) (-150) = 2948.3 m2 / s2 . Taking the square root gives: vy = +/- 54.30 m/s.
Remember that the square root can be positive or negative. In this case it's negative, because the y-component of the velocity will be directed down when the ball hits the ground.
Now we can use another equation to solve for time:
~
So, -54.30 = 2.873 - 9.8 t, which gives t = 5.834 seconds. Rounding off, the ball was in the air for 5.83 s.
We can use this time for part (b), to get the distance traveled in the x-direction during the course of its flight. The best equation to use is:
~
So, from the base of the cliff to the point of impact is 46.0 m.
A point about symmetry
At some point in its flight, the ball in the example above returned to the level of the top of the cliff (you threw it from the top of the cliff, it went up, and on its way down it passed through a point the same height off the ground as the top of the cliff). What was the ball's velocity when it passed this height? Its speed will be the same as the initial speed, 8.40 m/s, and its angle will be the same as the launch angle, only measured below the horizontal.
This is not just true of the initial height. At every height the ball passes through on the way up, there is a mirror-image point (at the same height, with the same speed, and the same angle, just down rather than up) on the downward part of the path.