

Projectile Motion
Name: Unknown
Status: N/A
Age: N/A
Location: N/A
Country: N/A
Date: Around 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.
Replies:
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 velocitycomponent 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 negativey direction (we have implicitly chosen
"up" as the positivey direction because of the +12 ycomponent 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 ydirection; 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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Update: June 2012

