kmarlow123 said:
A rocket shoots upward with a velocity of 500 feet/sec. Neglecting air resistance, how high will it travel? Is that even enough info to solve the problem? I don't remember how to solve that.
Well, if you assume that it left the launch pad at 500 feet/sec, and there's no thrust at that point, and assume constant acceleration from gravity, it's solvable. Not a very realistic scenario, but there you are.
Forgive me for converting to metric, but I despise doing anything with the imperial system. So let's say 500 feet/sec is about 152 meters/sec. You want the point where it starts to fall to earth. At that point, the velocity should be 0.
Gravity is about g = 9.81 m/s^2 and is constant. t is time, a is acceleration, and v is velocity.
-g = a = -9.81 m/s^2
Now take the integral with respect to t. You get v = -9.81t + C. At t = 0, we know v = 152 m/s, so 152 = -9.81(t) + C. So C = 152, giving an integral of v = -9.81t + 152.
Ok, now you need to know when it will be zero. So:
0 = -9.81t + 152
Solving for t, we get t = -152 / -9.81 = 15.5
So after about 15.5 seconds, it should start to fall back to earth.
Note that you can solve this pretty easily by thinking it through, however. I thought I'd show you how to solve this class of problem in general, but it shouldn't be hard to just think "Oh, 152 m/s, and it will lose 9.81 m/s every second, so to get to 0 velocity, it takes 152 / 9.81 = 15.5".
If you understood this, you can also tell me how long before it hits the ground, right? :) In fact, you should also know how to tell me how far up it went.
On a side note, this would be much more readable if I could get tex formatting working. It keeps showing some old equation for some reason.
