2.71
The quarterback Q throws the football when the receiver R is in the the position shown. The receiver’s
velocity is constant at 10 yd/sec, and he catches passes when the ball is 6 ft above the ground. If the
quarterback desires the receiver to catch the ball 2.5 sec after the launch instant shown, determine the initial
speed and angle θ required.
Given: Initial position and speed of receiver. Release point and time
Find: Speed and angle of throw
Assumptions: No acceleration in the x-direction, acceleration in the y-direction is equal to −g.
Solution:
a x t2
2
a y t2
y = yo + voy t +
2
xo = ax = 0
yo = 7 ft
ft
ay = −g = −32.2 2
s
ft
x = 90 ft + 30 (2.5 s) = 165 ft
s
y = 6 ft
voy
tan θ =
v
qox
2 + v2
vo = vox
oy
x = xo + vox t +
So,
x = xo + vox t
x − xo
165 ft
ft
vox =
=
= 66.0
t
2.5 s
s
gt2
y = yo + voy t −
2
gt
−1 ft 32.2 sft2 (2.5 s)
ft
y − yo
+
=
+
= 39.85
voy =
t
2
2.5s
2
s
!
39.85 fts
θ = arctan
= 0.543 rad (31.1◦ )
66.0 fts
s
2 2
ft
ft
ft
vo =
+ 39.85
= 77.1
66.0
s
s
s
2.81
The muzzle velocity of a long-range rifle at A is u = 400 m/s. Determine the two angles of elevation θ which
will permit the projectile to hit the mountain target B.
Given: y = 1500 m, x = 5000 m, vo = 400 m/s.
Find: θ
Assumptions: No acceleration in the x-direction, acceleration in the y-direction is equal to −g.
Solution:
a x t2
2
a y t2
y = yo + voy t +
2
xo = yo = ax = 0
x = xo + vox t +
ay = −g
vox = vo cos (θ)
voy = vo sin (θ)
Plugging in what we know,
x = vo cos (θ) t
(1)
2
y = vo sin (θ) t −
gt
2
(2)
We have two equations and two unknowns. Identifying that we need to solve these two equations is the
important part. The rest of the solution is just algebra. There are many ways to handle the algebra. Here
are two:
Solving (1) for t,
t=
x
vo cos θ
(3)
pluggining into (2),
g
h
x
vo cos θ
i2
x
−
vo cos θ
2
2
g
x
= x tan θ −
2 vo2 cos2 θ
g x2
sec2 θ
= x tan θ −
2 vo2
g x2
1 + tan2 θ
= x tan θ −
2 vo2
g x2
g x2
−y
0 = − 2 tan2 θ + x tan θ −
2 vo
2 vo2
y = vo sin (θ)
= A tan2 θ + B tan θ + C
√
−B ± B 2 − 4AC
tan θ =
2A
(4)
(5)
(6)
(7)
(8)
(9)
(10)
where,
A=−
9.81 sm2 (5000 m)2
gx2
=
−
= −766 m
2
2vo2
2 400 m
(11)
s
B = x = 5000 m
(12)
2
gx
− y = −766 m − 1500 m = −2266 m
2vo2
tan θ1 = 0.490 ⇒ θ1 = 0.456 rad (21.6◦ )
tan θ2 = 6.037 ⇒ θ2 = 1.41 rad (80.6◦)
C =−
(13)
(14)
(15)
Here is another way to do the algebra.
Solving (1) for cos θ,
cos θ =
x
vo t
(16)
Using cos2 θ + sin2 θ = 1 to replace sin θ in (2),
y = vo t
p
gt2
1 − cos2 θ −
2
(17)
y = vo t
s
(18)
plugging (15) into (18)
y=
1−
x
vo t
2
p
gt2
vo2 t2 − x2 −
2
−
gt2
2
(19)
rearranging
y+
squaring both sides
y 2 + ygt2 +
gt2 p 2 2
= t vo − x2
2
(20)
g2 4
t = t2 vo2 − x2
4
(21)
rearranging
g2 4
t + yg − vo2 t2 + x2 + y 2 = 0
4
2
A t2 + B t2 + C = 0
t2 =
(22)
−B ±
(23)
√
B 2 − 4AC
2A
(24)
with,
9.81 sm2
g2
=
A=
4
4
2
m2
= 24.1 4
s
m m 2
m2
B = yg − vo2 = (1500 m) 9.81 2 − 400
= −145285 2
s
s
s
2
2
2
2
2
C = x + y = (5000 m) + (1500 m) = 27250000 m
(25)
(26)
(27)
so,
2
t =
2
145285 m
s2
±
q
t2 = 193.8 s2 , 5845 s2
t1 = 13.9 s
2
−145285 m
s2
2
2
2
− 4 24.1 m
s4 27250000 m
m2
2 24.1 s4
t2 = 76.5 s
(28)
(29)
(30)
(31)
For each time use (1) to find θ.
θ = arccos
x
vo t
#
5000 m
= 0.456 rad (21.6◦ )
θ1 = arccos
(13.9
s)
400 m
s
"
#
5000 m
θ2 = arccos
= 1.41 rad (80.6◦ )
400 m
(76.5
s)
s
"
(32)
(33)
(34)
Height at a range of 5.0 km as a function of launch angle.
8000
7000
Height (m)
6000
5000
4000
3000
2000
1000
0
0
0.5
θ (radians)
1
1.5