- 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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Posted 15 September 1999 - 06:29 PM
THANKS.
Posted 15 September 1999 - 03:21 PM
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
Posted 15 September 1999 - 06:29 PM
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.
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-Kentamanos
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