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Guest Anonymous Poster

Trajectory of a ball

2 posts in this topic

here's the formula i put on my calculator last year so that i could remember it on math test =) but then i sorta entered stuff over it, so here's my guess:

height = -.5g(t*t) + v*t + h

g gravity constant, i think it's 9.8 for normal gravity

t time
v initial velocity
h initial height

i hope that helps

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I need the equation for trajectory
- 30 degree up and 20 m/s . Now I want to calculate the x and y Position on a certain Time .
Or the xmove and ymove on a certain Time . Any useful hints ?


THANKS.

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For simplicity sake, I will call this problem a cannon with a cannon ball. You might have something different, but I'm sure you'll know what I mean. If you'd like, just skip down to the bottom to get the equations, but you might want to know why these are the equations.

Physics books will tell you that the linear position for an object at a given time with a constant acceleration is:

s = s0 + v0*t + (1/2)*a*t^2

where s0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is time.

For your example (basically a simple 2D cannon), you have to break it up into vertical and horizontal components.

To start with, you need an initial velocity for the horizontal component (vx0) and an initial velocity of the vertical component (vy0). To do that you have to use a little trig. If you are saying the the cannon has a 30 degree inclination, the values of these would be like so:

vx0 = cos(30 degrees)*20
vy0 = sin(30 degrees)*20

You also need to take into consideration the acceleration on the cannon ball. The only acceleration component is in the vertical direction, gravity. Gravity is usually considered to be 9.81 m/s in the negative (downward) direction. Since there is no acceleration in the horizontal direction, this simplifies the equation for x.

This leaves us with the equations (in metric units):
x = x0 + cos(30)*20*t
y = y0 + sin(30)*20*t - (1/2)(9.81)*t^2

Make sure your coordinate system is the correct orientation and the initial velocities are the correct sign.

If you have any questions, let me know. My physics knowledge is a little rusty, but these should be correct. If you want to discuss the calculus behind this, let me know too...hehehe.

------------------

-Kentamanos

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