Range and Height of Projectiles
To solve the equations for projectile
problems one needs to know the initial conditions for the position
(x0,y0) and velocity
(vx0,vy0) at the initial time (t0 =
0). Commonly one sets x0=y0=0 and gives the
initial velocity in terms of its absolute value, v0, and
angle to the x-axis, $\theta$0, i.e.
in polar coordinates:
v0 =
(vx02+vy02)1/2;
\[ \rm $\theta$_{0} = tan^{-1}\frac{v_{y0}}{v_{x0}} \]
v0x = v0 cos$\theta$0;
v0y = v0 sin$\theta$0
We can now ask the following questions:
- Question: Supposing that the starting and
finishing height (y-coordinate) of a projectile are the same, what
is the distance covered in x-direction? [This is called the
'range', R, of the
projectile].
Answer:
x - x
0 = [v02/g]sin(2$\theta$)
- Question: What is the
maximum height reached during
the trajectory?
Answer:
h
max = [v02/2g]sin2($\theta$)
- Question: What is the
shape of the trajectory in the
x-y plane?
Answer:
It is a parabola,
\[ \rm y(x) = xtan$\theta$_{0} - \frac{x^{2}g}{2v_{0}^{2}cos^{2}$\theta$_{0}} \]
© MultiMedia Physics, 1999