Physics - Projectile Motion Problems
One of the most mind boggling parts of physics is the very first section - involving projectile motion and the 4 kinematics equations. These equations assume that there is constant acceleration, given by the acceleration of gravity which is the g constant you see in the formulas (9.8 m/s^2).
When an object moves through the air (and ignoring the effects of air resistance), it moves in two dimensions - an X and a Y component. The first step is to separate the X and Y, meaning - you can apply those 4 formulas to the X direction (horizontally) and you can also apply them in the Y-direction (vertically).
Imagine if you throw a ball sideways off the cliff that is 3m high, and you want to know how long it takes.
Since both of the x and y parts are happening at the same time, that means that the Time that you use for both equations is the same. This means conceptually - that if I shot something straight sideways across a field with, let's say, a cannon, it would take the same amount of time to reach the ground as if I had simply dropped the cannonball instead. Weird, isn't it? The reason is that gravity does not affect the horizontal part of the motion, only the vertical part.
So the first step is to look at all the equations and pick the one that has most of the elements you have, but is missing the elements you are looking for.
X=VoT + 1/2AT^2 is a good choice because you have the height and you want the time.
In the vertical Y-direction you can just rewrite this (replacing the X with Y, and replacing the A with g): Y= VoT + 1/2(g)T^2.
First thing to note is that in the Y-direction (up and down), there is no Initial Velocity. You did not throw the ball straight UP and you did not push it DOWN when it left your hand, you threw it SIDEWAYS, right? Thus Vo=0 in the vertical direction and the whole VoT becomes equal to 0.
You are left with Y=1/2gT^2 and this is where the equation
Height(h)= 1/2gt^2 comes from.
So here goes:
3 = 1/2gT^2
3 = 0 + (1/2)(9.8)T^2
T^2 = 2*3/9.8 = 0.6122
T = 0.78 sec
H=1/2gt^2 can be used in situations where things are flying sideways off cliffs or when simply dropped, like from an airplane..where you did not apply an acceleration to it by lobbing it UP in the air or pushing it DOWN.
Remember this equation: (H=1/2gt^2) it will come in handy!
And best of luck on your physics exams! Send me an email and let me clarify this and other mind-boggling physics concepts to you or your student! I would say "It's not rocket science!" but the scientists at NASA probably DO use these equations ;-)