Setting Up a Basic Projectile Problem

Physics Tutorial

Setting Up a Basic Projectile Problem

Intro

In physics, the crux of any problem is finding what is unknown and then knowing it. In most basic projectile problems, these boil down to velocity, time, and displacement.

Sample Problem

A ball is thrown upwards with a velocity of 10 m/s. How high does it go and how long does it take to get there? Assume the acceleration due to gravity is g = 10 m/s^2

A ball reaches height h from initial velocity Vo

Solution

In a one-dimensional projectile problem there are really only five things to know. Initial velocity, final velocity, time taken, distance traveled, and the acceleration of the object, and if you know any three, you can find out the other two.

So in this case, we were told our initial velocity Vo = 10 m/s, our acceleration of the object is just gravity so g= 10 m/s^2, and we know that the ball has reached its highest when it stops moving upward, so we know our final velocity Vf = 0m/s.

That leaves height h and time t.

Vo= 10 m/s
Vf= 0 m/s
a=10 m/s^2
t=?
h=?

The definition of acceleration is change in velocity over change in time, or rather
a = (Vo-Vf)/t
and we know the change in velocity is 10 m/s and a = 10 m/s^2 so
10 = 10/t
and a little algebra shows
t = 1 s
so we now know it takes just one second for this ball to get to the highest point!

Now we want to know h. The definition of average velocity Vavg is total distance traveled divided by total time taken, or
Vavg = h/t .
But the average velocity is also the statistical average of the initial and final velocities (because the acceleration is constant), or rather,
Vavg = (Vf-Vo)/2
In this case Vavg is 5 m/s.
Then we remember
Vavg = h/t
5 = h/1
5 = h
and now we know all five things.

Vo= 10 m/s
Vf= 0 m/s
a=10 m/s^2
t= 1s
h= 5m



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