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Projectile Equation
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Question:
I am writing a computer program, and do not remember what the
equation to describe a projectile in motion. I need to program the computer so
that my basketball player will shoot accurately with respect to gravity. I
thought the equation was a parabola, but that does not seem quite right.
Can anyone help me?
Replies:
Yup, it should be a parabola. THe equation for projectile
motion is the same for any projectile - you can look it
up in any mechanics text, but here it is again anyway:
x(t) = x0 + t v0 + t^2/2 a0
where everything except t is a vector, t is the time, x
is position, x0 is initial position, v0 is initial
velocity, and a0 is the (constant) acceleration of gravity
in the downward direction at 9.8 m/s^2
Arthur Smith
I do not think that is quite what Dave needs. The motions in the
horizontal(here denoted by x) and vertical (here denoted by y) directions, as
functions of time, are:
x = x0 + V*cos(theta)*t and y = y0 + V*sin(theta)*t - (1/2)*g*t^2
where (x0, y0) is the location of the ball, relative to some arbitrarily
chosen origin of coordinates, at time t=0 (the time when the ball leaves
the player's hands), V is the speed of the ball at time t=0, theta is the
angle that the velocity vector makes with the horizontal at time t=0, and
g is the acceleration due to gravity (g = 9.8 m/sec^2 if x and y are in
meters, positive y is "up", and t is in seconds.) You *can* solve the x-
equation for t and substitute that into the y-equation to show that the
path taken by the basketball (ignoring air resistance and other air-
related effects) is a parabola. So you could choose x-values, compute the
corresponding y-values, and plot these pairs. You could also choose values
for t and compute x and y at these different times using the x and y equa-
tions above.
Rcwinther
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Update: June 2012
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