Physics

200512112135Bubba

Bubba, the village idiot, shoots an arrow straight up into the air at a velocity of 100.0 meters per second. Answer the following questions.

A) If he does not move, how long does he have to live?

B) What is the maximum height that the arrow will reach?

C) Where will the arrow be after 7.000 seconds?

D) What is the velocity of the arrow after 13 seconds?

E) After 10 seconds Bubba wises up and decides that he should move so he dives under his Ford F-150 pickup truck, which is parked close by. Where will the arrow be 30.00 seconds after he shot it into the air?

Other information: For this problem neglect air resistance.
Assume that the acceleration due to gravity is the usual 9.800 m/s^{2}.
Also assume that there is no wind and that the arrow goes up and down along the
exact same path. Also Bubba releases the arrow 2.000 meters above the
ground since he is about 2 meters tall. All answers are to be accurate to
4 significant digits. Show your equations and your work. Think about
the directions of the various quantities.

Solution for A:

Question: How long does Bubba have left to live?

If Bubba does not move the arrow will go straight up, turn around, then fall back toward Bubba's head. Since we assume perfect conditions the arrow will return to the release point at a velocity of -100.0 meters/second since we know that g is constant.

Here is what we know:

Up is positive, down is negative.

v_{i }= +100.0 meters/second since the arrow was shot upward.

v_{f} = -100.0 meters/second since the arrow is returning toward Bubba's
head.

a = g = -9.800 meters/second^{2} since gravity pulls objects down.

t = ? (This is what we want to determine in part A)

Use: v_{f }= v_{i} + at (Solve the formula for time t)

t = (v_{f}-v_{i})/a

t = ((-100.0 m/s) - (100 m/s))/(-9.800 m/s^{2})

t = (-200.0 m/s )/(-9.800 m/s^{2})

t = (200.0/9.800) s

t = 20.40816327 s (Round to 4 significant digits)

**t = 20.41 seconds**

If Bubba does not move he has 20.41 seconds left to continue life as a
biological entity after he releases the arrow straight up into the air.

Since Bubba, after 10 seconds, does realize that he should move, we can assume that there might be some hope for Bubba to become at least somewhat less dangerous to himself and to others.

Solution for B:

Question: What's the maximum height the arrow reaches?

Once released from the bow by Bubba, the arrow will be slowed by gravity until it stops going up. At the point where it stops going up its upward velocity will be 0.000 meters/second. Here we will calculate how high the arrow goes above the release point, which is the same place as Bubba's head. This vertical displacement, ∆y, will be positive since it is in an upward direction. Remember that we chose upward to be positive.

Here is what we know:

Up is positive, down is negative.

v_{i }= +100.0 meters/second since the arrow was shot upward.

v_{f} = 0.000 meters/second since the arrow must stop at its highest
location

a = g = -9.800 meters/second^{2} since gravity pulls objects down.

∆y = ? (This is
what we want to determine in part B)

Use the equation: v_{f}^{2} = v_{i}^{2} + 2a∆y
(Solve for ∆y)

∆y = ( v_{f}^{2}
- v_{i}^{2} )/2a (Since vf is 0 we can eliminate it from the
equation)

∆y = ( - v_{i}^{2}
)/2a

∆y = ( - (100m/s)^{2}
)/2(-9.800 m/s^{2})

∆y = ( -10000m^{2}/s^{2}
)/(-19.60 m/s^{2}) (Negative divided by negative is positive)

∆y = (10000m^{2}/s^{2}
)/(19.60 m/s^{2})

∆y = 510.2040816 m (Round
to 4 significant digits)

**∆y = 510.2 m**

The arrow reaches its maximum height of 510.2 meters above the release point which was at Bubba's head. Since Bubba's head is 2.000 meters above the ground the arrow will be 510.2 meters + 2.0 meters, or 512.2 meters AGL. AGL means above ground level. You will probably need to know this acronym sometime in the not to distant future.

Solution for C:

Question: Where will the arrow be after 7.000 seconds?

For calculation purposes we will use the starting point as being the location where Bubba released the arrow.

Here is what we know:

Up is positive, down is negative.

v_{i }= +100.0 meters/second since the arrow was shot upward.

a = g = -9.800 meters/second^{2} since gravity pulls objects down.

t = 7.000 s (This is only for this section)

∆y = ? (This is
what we want to determine in part C, but notice that this is a different ∆y
than it was in the previous section. This is the ∆y after only 7.000
seconds of flight.)

Use the formula: ∆y = v_{i}t
+ ^{1}/_{2}at^{2}

∆y = v_{i}t + ^{1}/_{2}at^{2}

∆y = (100 m/s)(7 s) + ( ^{1}/_{2})(-9.800 m/s^{2})(7.000
s)^{2}

∆y = (700 m) + (-4.900 m/s^{2})(49.000 s^{2})

∆y = (700 m) + (-240.1 m)

**∆y = 459.9 m **

After 7.000 seconds the arrow will be 459.9 meters above the release point. It will be 459.9 meters + 2.0 meters or 461.9 meters AGL.

Solution for D:

Question: What is the velocity of the arrow after 13 seconds?

Here is what we know:

Up is positive, down is negative.

v_{i }= +100.0 meters/second since the arrow was shot upward.

a = g = -9.800 meters/second^{2} since gravity pulls objects down.

t = 13.000 s (This is only for this section)

vf = ? (This is what we are looking to determine. This will be the
velocity after 13.00 seconds.

Use the equation: v_{f }= v_{i} + at

v_{f }= v_{i} + at

v_{f }= (100 m/s) + (-9.800 m/s^{2})(13.00 s)

v_{f }= (100 m/s) + (-127.4 m/s)

**v _{f }= -27.4 m/s **

Since this velocity is negative, and since we chose down to be negative, the arrow is returning toward the location where Bubba's head was only 3 seconds ago. Once again, remember to pay close attention to the + or - signs when doing physics problems. They are very important.

Since for this section we only wanted the velocity we do not need to do anything with the fact that Bubba released the arrow 2 meters off the ground.

Solution for E:

Question: Where is the arrow after 30 seconds from its release?

For calculation purposes we will use the starting point as being the location where Bubba released the arrow.

Here is what we know:

Up is positive, down is negative.

v_{i }= +100.0 meters/second since the arrow was shot upward.

a = g = -9.800 meters/second^{2} since gravity pulls objects down.

t = 30.000 s (This is only for this section)

∆y = ? (This is
what we want to determine in part E, but notice that this is a different ∆y
than it was in the section B and in section C. This is the ∆y after
30.00 seconds of flight.)

Use the formula: ∆y = v_{i}t
+ ^{1}/_{2}at^{2}

∆y = v_{i}t + ^{1}/_{2}at^{2}

∆y = (100 m/s)(30 s) + ( ^{1}/_{2})(-9.800 m/s^{2})(30.000
s)^{2}

∆y = (3000 m) + (-4.900 m/s^{2})(900.000 s^{2})

∆y = (3000 m) + (-4410 m)

**∆y = -1410 m (Correct mathematically - but not a reasonable answer)**

Mathematically the arrow will be 1410 meters below the release point or 1408 meters into the ground. However, in the real world of physics, this is not even remotely logical. A falling arrow will not likely bury itself 1408 meters into the ground. The arrow will be stuck in the ground where Bubba had been standing before he decided he should hide under his truck.

Common sense, which is not all that common, tells us that since the arrow had already returned to the release point after 20.41 seconds it could not continue for the rest of the 30 seconds. It must be stuck in the ground. This could have been determined without any additional calculation.

Bubba's decision to move allows him to live on for future adventures in his, and your, study of physics.