Transcript Notes

Projectile Motion Example Problem 1
A player shoots a free throw to a basket 15 feet away and 10 feet
off the floor. If the ball is released from a point 7 feet above the
floor and at an angle of 50º, determine:
(1) The required initial velocity, v0;
(2) The time the ball passes through the rim;
(3) The maximum height of the trajectory;
(4) The speed of the ball and the angle of
its trajectory as it passes
through the rim.
50°
10 ft
7 ft
15 ft
First establish an x-y coordinate
system that makes sense for the
problem:
y
50°
10 ft
x
7 ft
y
15 ft
v0 = ?
50°
10 ft
7 ft
15 ft
x
Write the two projectile position equations:
x = x0 + (v0·cos q)·t
15 = 0 + (v0·cos 50)·t
15 = .643·v0·t
y = y0 + (v0·sin q)·t – ½gt2
10 = 7 + (v0·sin 50)·t – ½(32.2)t2
3 = .766·v0·t – 16.1·t2
y
It’s easy to solve
these two equations
(see the next page)
for the two unknowns
(v0 and t).
v0 = ?
50°
10 ft
7 ft
15 ft
x
There are several ways to solve for v0 and t, but I prefer to
substitute the v0t product from the first equation into the
second:
15 = .643·v0·t
3 = .766·v0·t – 16.1·t2
23.34 = v0·t
3 = .766·(23.34)– 16.1·t2
16.1·t2 = 17.88 – 3 =
14.88
t = 0.961 sec
23.34 = v0·t
v0 = 24.3 fps
Now find the height of the apex of the trajectory:
v0 = 24.3 fps
Easiest equation to use for this:
v0y = 24.3(sin 50) = 18.6 fps
vy = 0 at apex
y
vy2 = v0y2 – 2g(y – y0)
0 = (18.6)2 – 2(32.2)(y - 7)
y = hmax = 12.38 ft
v0 = ?
v0y
50°
v0x
7 ft
hmax = ?
10 ft
(x0, y0)
= (0,7) ft
15 ft
x
This is a low
trajectory. A good
free throw should
have a larger launch
angle and a
higher arc.
Finally, find the speed and angle of the ball
as it passes through the rim:
vx = 24.3 cos 50 = 15.62 fps
v
vy = v0sin 50 - gt
= 24.3 sin 50 - 32.2(.961)
= - 12.33 fps = 12.33 fps
v = 15.62 + 12.32
= 19.9 fps
= tan-1
12.33
19.9
= 38.3°