Projectile Motion

(Created prior to 1993)

Question: If a football is kicked with velocity vector:
v = (16m/s)x + (12m/s)y  then how can I calculate: 
1) the speed with which it hits the ground, 
2) the angle it makes hitting the ground, 
3) its "hang time"  
4) it range and 
5) its maximum height?  Basically, what equations are needed here, concepts, 
etc.
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For convenience, let us assume that the ball's location prior to 
being kicked is x=0, y=0, and that it is kicked at time t=0.  Also let us 
ignore air friction.  Then there is no horizontal force acting on the ball, 
and gravity is the only vertical force.  The velocity-component equations are:  
v_x(t) = 16, v_y(t) = 12 - 9.8*t [both come from v(t) = v0 + a*t]; note that 
the acceleration due to gravity, 9.8 m/s^2, is negative in the v_y equation 
because gravity acts in the negative-y direction (we have implicitly chosen 
"up" as the positive-y direction because of the +12 y-component of the initial 
velocity).  The position equations are:  x(t) = 16*t, y(t) = 12*t - 4.9*t^2 
[both come from s(t) = s0 + v0*t + (a*t^2)/2 ].  The ball hits the ground when 
y=0; setting y(t)=0 and solving, we get t=0 or t=12/4.9 ~ 2.449 s.  The later 
time is the answer to (3).  To answer (1), evaluate the v_x(t) and v_y(t) 
equations at t=12/4.9, then use the formula speed = sqrt[(v_x)^2 + (v_y)^2] = 
sqrt[16^2 + (-12)^2]= 20 m/s.  Draw a picture of the x and y components of 
velocity and their vector sum at impact, with all three vector tails at a 
common point, and use trig to find the answer to (2) (its arctan(-12/16), or ~ 
-36.87 degrees).  To get the range, use the x(t) equation evaluated at the 
time of impact:  range = 16*12/4.9 ~ 39.18 m.  Finally, to get the maximum 
height, recognize that this occurs at the time when the ball momentarily has 
zero velocity in the y-direction; set v_y(t)=0 and solve for t.  This gives 
t=12/9.8 ~ 1.224 s.  Evaluate y(t) at t=12/9.8 to get the answer to (5):  max 
height = 72/9.8 ~ 7.347 m.

R.C. Winther
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