Answers and Selected Solutions to Vectors 3 worksheet
1.
A baseball is thrown from the stands 32 ft above the field at an angle of 36°. When and how far away will the ball strike the
ground if its initial speed is 32 ft/sec?
time ≈ 2.119 seconds; distance ≈ 54.87 feet
2.
What two angles of elevation will enable a projectile to reach a target 16 km away at the same elevation as the gun that fired the
projectile, if the projectile's initial speed is 400 m/sec?
!
!
!
!
!
a ( t ) = −9.8 j ⇒ v ( t ) = 400cos (θ ) i + 400sin (θ ) − 9.8t j
!
!
!
⇒ r ( t ) = t ⋅ 400cos (θ ) i + t ⋅ 400sin (θ ) − 4.9t 2 j
(
(
) (
) (
t ⋅ 400cos (θ ) = 16000 ⇒ t =
)
)
40
cos (θ )
2
⎛ 40 ⎞
⎛ 40 ⎞
t ⋅ 400sin (θ ) − 4.9t 2 = 0 ⇒ ⎜
⋅
400sin
θ
−
4.9
(
)
⎟
⎜
⎟ =0
⎝ cos (θ ) ⎠
⎝ cos (θ ) ⎠
(
)
(
⇒ θ ≈ 39.3° 0.685r or θ ≈ 50.7° 0.886r
)
3.
An electron in a TV tube is beamed horizontally at a speed of 5 × 106 m/sec toward the face of the tube 40 cm away. About how
far will the electron drop before it hits the tube?
drop ≈ 3.136 × 10−14 m or 3.136 × 10−12 cm
4.
Laboratory tests designed to determine how far golf balls of different hardness travel when hit with a driver showed that a certain
golf ball hit with a club-head speed of 100 mph at a launch angle of 9° carried 248.8 yards. What was the launch speed of the
ball (and no, it was NOT 100 mph)?
!
!
!
!
!
a ( t ) = −32 j ⇒ v ( t ) = v0 cos ( 9° ) i + v0 sin ( 9° ) − 32t j
!
!
!
⇒ r ( t ) = t ⋅ v0 cos ( 9° ) i + t ⋅ v0 sin ( 9° ) − 16t 2 j
(
(
) (
) (
)
)
t ⋅ v0 cos ( 9° ) = 248.8 × 3 ⇒ t ≈
746.4
v0 cos ( 9° )
( Must convert yards to feet since units for acceleration are ft/sec.)
2
⎛ 746.4 ⎞
⎛ 746.4 ⎞
t ⋅ v0 sin ( 9° ) − 16t 2 = 0 ⇒ ⎜
⋅
v
sin
9°
−
16
(
)
⎟ 0
⎜
⎟ =0
⎝ v0 cos ( 9° ) ⎠
⎝ v0 cos ( 9° ) ⎠
⇒ v0 ≈ 278.016 ft/sec or 189.556 mph
5.
Show that a projectile attains three-quarters of its maximum height in half the time it takes to reach its maximum height.
!
!
!
!
!
a ( t ) = −gj ⇒ v ( t ) = v0 cos (θ ) i + v0 sin (θ ) − gt j
(
) (
)
! ⎛
⎞!
1
!
⇒ r ( t ) = ( t ⋅ v cos (θ )) i + ⎜ t ⋅ v sin (θ ) − gt ⎟ j
2
⎝
⎠
2
0
0
!
Projectile will reach its maximum height when v ( t ) =0.
v0 sin (θ ) − gt = 0 ⇒ t =
v0 sin (θ )
g
1
y ( t ) = t ⋅ v0 sin (θ ) − gt 2
2
⎛ v sin (θ ) ⎞
Maximum height = y ⎜ 0
⎟
g
⎝
⎠
⎛ v sin (θ ) ⎞
1 ⎛ v0 sin (θ ) ⎞
=⎜ 0
⋅
v
sin
θ
−
g
(
)
⎟ 0
⎟
g
2 ⎜⎝
g
⎝
⎠
⎠
=
=
v0 2 sin 2 (θ )
g
−
v0 2 sin 2 (θ )
2g
v0 2 sin 2 (θ )
2g
⎛ 1 v sin (θ ) ⎞ ⎛ v0 sin (θ ) ⎞
1 ⎛ v0 sin (θ ) ⎞
y⎜ ⋅ 0
=
⋅
v
sin
θ
−
g
(
)
⎟ ⎜ 2g ⎟ 0
g
2 ⎜⎝ 2g ⎟⎠
⎝2
⎠ ⎝
⎠
=
v0 2 sin 2 (θ )
2g
−
2
v0 2 sin 2 (θ )
3 v0 sin (θ )
8
g
2
=
2
8g
2
2
2
3 v0 sin (θ )
= ⋅
4
2g
6.
=
3 ⎛ v0 sin (θ ) ⎞
y
⎟
4 ⎜⎝
g
⎠
=
3
( Maximum height )
4
A golf ball leaves the ground at a 30° angle at a speed of 90 ft/sec. Will it clear the top of a 30-ft tree 135 ft away?
Golf ball is about 29.994 feet above the ground when it has traveled 135 feet horizontally
and reaches a height of 30 feet when it has traveled about 135.56 feet horizontally.
From either perspective, the golf ball does not quite clear the tree.
7.
A person who throws a baseball at 96 feet per second (about 65 miles per hour) is standing 24 feet from the base of a building. The ball just
barely makes it onto the roof. How tall is the building? (Note: the answer you compute should not actually make sense—the building should
be about fifteen stories tall, and no one can really throw a ball onto a fifteen-story building. Can you explain what went wrong?)
Let’s say the angle of inclination of the throw was θ. Then the horizontal velocity of the ball is 96cos(θ), and the time needed to travel the 24
horizontal feet to the building is 24/(96cos(θ)) = 1/(4cos(θ)). The vertical height of the ball at time t is given by –16t2 + v0t (initial height being
zero in this case). The initial vertical speed is v0 = 96sin(θ). Plugging in this and the time t = 1/(4cos(θ)) gives a vertical height of –sec2(θ) +
24tan(θ).
Let’s find the θ that maximizes this. Differentiating with respect to θ gives h′(θ) = 24sec2(θ) – 2sec2(θ)tan(θ). Setting this equal to zero
requires either sec2(θ) = 0 (impossible) or tan(θ) = 12, or about 85.24°. The associated height is 143 feet (exactly!).
This is the height of a fourteen- or fifteen-story building, so it doesn’t make any sense in context. No one can throw a baseball that high. What
went wrong? Well, when a baseball pitcher throws he or she is throwing horizontally or even a little downward. Throwing upward is not
natural, and humans cannot throw upward with nearly the velocity that they can throw horizontally. A good near-vertical throw might be at half
the speed (or less) of a good horizontal throw. Half the speed gets a throw of ¼ the distance. Or less. A fifteen-story building is ridiculous, but
¼ of that, or a 3-4 story building, isn’t too outrageous to throw to the roof.