1. A roadrunner runs directly off a cliff with an initial velocity of 3.5 m/s
a) What
are the components of this velocity?
Vx = 3.5 m/s Vy = 0 m/s
b) What
will be the horizontal velocity 2 seconds after the bird leaves the cliff?
3.5 m/s – horizontal velocity is unchanging
c) If
the cliff is 300 m high, at what time will the roadrunner reach the ground?
h = dy = ½ x 10 x t2 =
300
300 x 2 / 10 = t2 = 60
t = 7.75 s
d) How
far from the cliff will this bird land?
dx = 3.5 x 7.75 = 27.125 m
e) If
there is a small pond which begins 25m away from the cliff and extends 2.5
meters from there; will the roadrunner land in the pond?
Yes, the pond is
from 25 m to 27.5 m, so the roadrunner will land in the pond.
f) What
is the final vertical velocity at which the roadrunner is traveling? [The
vertical velocity at the time when the bird reaches the ground]
Vy = 10 x 7.75 + 0 = 77.5 m/s
g) What
is the final horizontal velocity at which the roadrunner is traveling? [The
horizontal velocity at the time when the bird reaches the ground]
Vx = 0
+ 3.5 = 3.5 m/s
h) What
is the total final velocity of this motion? [magnitude and direction]
V2 = 77.52 + 3.52
= 6018.5
V = 77.579 m/s
q = tan-1 (77.5 / 3.5) = 87.41o below
the horizontal
2. An object (any object) is dropped from a height of 300m
a) How
long does it take this object to fall to the ground?
h = dy = ½ x 10 x t2 =
300
300 x 2 / 10 = t2 = 60
t = 7.75 s
b) Compare
this answer with you answer from question 1, part c). What are the reasons for
any similarities or differences?
They are the same because their
vertical motions are identical. All objects fall with the same gravitational
acceleration, so two objects at the same height with the same initial vertical
velocity will reach the ground at the same time.
3. The
intent of a bean bag toss game is to get your bean bag to land on the
‘bull’s-eye’ of a target. The target is set up parallel to the ground and is
the same height above the ground as your hand is when you let go of the bean
bag. The game’s rules further require you to be 5 m from the center of the
target when you release the bag.
a) Evaluate
the following questions for both an angle of 32o and an angle of 58o
if the bean bag is thrown with an initial velocity of 6 m/s:
32o 58o
i.
What are the components of velocity?
Vx32:
Cos 32o = Vx32/6 Vx58: Cos 58o = Vx58/6
6 x Cos 32o
= Vx32 = 5.09 m/s 6 x Cos 58o = Vx58 = 3.18 m/s
Vy32:
Sin 32o = Vy32/6 Vy58:
Sin 58o = Vy58/6
6 x Sin 32o
= Vy32 = 3.18 m/s 6 x Sin 58o = Vy58 = 5.09 m/s
ii. What
is the maximum height of the bean bag’s motion?
tTOP32
= Vy/10 = 3.18/10 = 0.318s tTOP58
= Vy58/10 = 5.09/10 = 0.509s
hMAX32
= ½ x 10 x 0.3182 = 0.506 m hMAX58
= ½ x 10 x 0.5092 = 1.295 m
iii. How
long will the bean bag be in the air?
tTOTAL32
= 2 x tTOP32 = 0.636 s tTOTAL58
= 2 x tTOP58 = 1.018 s
iv. How
far away from you will the bag land?
dx32
= 5.09 x 0.636 = 3.24 m dx58 = 3.18 x 1.018 = 3.24m
v. If
the center of the bull’s-eye ranges from 4.9 m to 5.1 m away from you, does
your bean bag win?
No No
4. A
stunt driver drives a red mustang convertible up a ramp and off a cliff. The
car leaves the ramp at a velocity of 60 m/s at an angle of 45o to
the horizontal; the cliff and ramp combined cause the car to begin its
projectile motion at a height of 315m above the ground. If you were
coordinating this stunt, how far away would you put a landing surface so that
your stunt driver was not injured?
In order to
find horizontal distance we need horizontal velocity and time. We can find both
horizontal and vertical velocity from the initial conditions, but we’ll have to
calculate the time it will take for the car to reach the ground. So first we’ll
find the components of velocity:
Vx:
Cos 45o = Vx/60 Vy:
Sin 45o = Vy/60
60 x Cos 45o
= Vx = 42.43 m/s 60 x Sin 45o = Vy = 42.43 m/s
Horizontal velocity is constant, but the
vertical velocity we calculated above is only the initial vertical velocity.
We use the initial vertical
velocity to find the time it takes the car to reach the top of its path and
fall to the ground. Let’s think about this in two parts; the time it takes to
reach hMAX first:
tTOP = 42.43/10 = 4.243 s
Now what about the time it takes to fall from
the maximum height? Well first we need to know the maximum height:
hTOP = ½ x 10 x 4.2432 =
90.02 m
hMAX = hTOP + ho
= 90.02 + 315 = 405.02 m
Now we calculate the time it takes to fall
from a height of 405.02 m:
405.02 = ½ x 10 x tDOWN2
tDOWN2 = 81.004
tDOWN = 9.000 s
Putting these two times together, we have the
total time it takes the car to travel up to its maximum height and then fall
back down. This is the total time in the air and this is the time we will want
to use to solve for horizontal distance.
tTOTAL = tTOP + tDOWN
= 4.243 + 9.000 = 13.243 s
dx = 42.43 x 13.243 = 561.9 m
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