Upside down falling body problem

Click For Summary
SUMMARY

The discussion centers on calculating the maximum height a rocket reaches when launched upward at a velocity of 500 feet/sec (approximately 152 meters/sec) while neglecting air resistance. Using the constant acceleration of gravity (g = 9.81 m/s²), the time to reach the peak height is determined to be 15.5 seconds, at which point the velocity becomes zero. The problem-solving approach involves integrating the velocity function and solving for time, demonstrating a straightforward method to analyze similar projectile motion scenarios.

PREREQUISITES
  • Understanding of basic physics concepts, specifically kinematics.
  • Familiarity with calculus, particularly integration and solving equations.
  • Knowledge of gravitational acceleration (g = 9.81 m/s²).
  • Ability to convert between imperial and metric units.
NEXT STEPS
  • Learn more about projectile motion equations in physics.
  • Study the effects of air resistance on projectile trajectories.
  • Explore advanced kinematics involving multiple forces acting on a body.
  • Investigate numerical methods for solving differential equations in motion analysis.
USEFUL FOR

Students of physics, engineers, and anyone interested in understanding the principles of motion and projectile dynamics.

kmarlow123
Messages
4
Reaction score
0
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.
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
18
Views
3K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 60 ·
3
Replies
60
Views
6K