How Far III: | 1 | 2 |
3 | 4 | 5 | 6 |
7 | 8 | 9
| 10 | Go up

** - by Chris Murray, 2002**

**1. Helen Wheels launches a rocket in the air for a total
of 7.48 seconds. What time did it spend going up? What is its initial velocity?
What is its final velocity? How high did it go? What is the rocket's velocity at
an elevation of 50.0 m? (Why are there two answers?) **

Let's go from the ground to the top. This is what we know:

Dx =

v_{i}=

v_{f}= 0 m/s (It is at the top)

a = -9.8 m/s/s (gravity)

Dt = 7.48/2 = 3.74 s (It takes half the time to reach the top)

So it spent 3.74 s going up, and we can find the initial velocity using:

v_{f}= v_{i}+ a(Dt)

0 = v_{i}+ (-9.8 m/s/s)(3.74 s)

v_{i}= 36.652 m/s = 36.7 m/s

The final velocity would just be negative this as it is going down, so just before it strikes the ground, it is going -36.7 m/s

Now you can find out how high it goes using

Dx =^{1}/_{2}(v_{i}+ v_{f})Dt

Dx =^{1}/_{2}(36.652 m/s + 0)(3.74 s) = 68.53924 m = 68.5 m

For the elevation of 50.0 m, we are going to need to start over. We know:

Dx = 50.0 m (given)

v_{i}= 36.652 m/s (from previous part)

v_{f}= ??? (It's not at the top, so we don't know what it is)

a = -9.8 m/s/s (gravity)

Dt = ??? (Again, we are not at the top, so we don't know,andwe don't care)

Use:

v_{f}^{2}= v_{i}^{2}+ 2a(Dx)

|v_{f}| = Ö^{}{v_{i}^{2}+ 2a(Dx)} = Ö{(36.652 m/s)^{2}+ 2(-9.8 m/s/s)(50.0 m)} = 19.1 m/s, so the answer could be +19.1 m/s, or -19.1 m/s, the first on the way up, and the second on the way down, as it passes an elevation of 50.0 m both times on its way up to the top.

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**2. Honda driven by Cliff Jumper going 27 m/s is 2.1
kilometers (km) behind another Pinto driven by Ima Sloe going 23 m/s in the same
direction. What time will it take the Cliff to overtake Ima? What distance will
the Cliff travel in this time? **

If you were standing on top of Ima's car looking at Cliff's, you would see him approaching you at 27 m/s - 24 m/s = 4 m/s, and you would also see him 2100 m behind you, and so we can figure out the overtaking time by simply using:

v_{avg}= Dx/Dt

(4 m/s) = (2100m)/Dt

Dt = 525 s

Cliff would have traveled:

v_{avg}= Dx/Dt

27 m/s = Dx/(525 s)

Dx = 14,175 m or about 14.2 km

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**3. Freda Dadark throws a ball up from the roof of a
building at a height of 21.7 m. It strikes the ground 3.78 seconds later. What
is the initial velocity of the ball? With what velocity does the ball hit the
ground? How high above the building does the ball go? What is the velocity and
displacement 2.12 seconds after the ball was released? **

From the top of the building to the ground, we have

Dx = -21.7 m (Down is negative)

v_{i}= ??

v_{f}= ??

a = -9.8 m/s/s (gravity)

Dt = 3.78 s

Even though the ball goes up, and then down, we can use this formula, as Dx is the displacement, not the distance. displacement is the net change in position, and this change in position is down or -21.7 m:

Dx = v_{i}(Dt) +^{1}/_{2}a(Dt)^{2}

(-21.7 m) = v_{i}(3.78 s) +^{1}/_{2}(-9.8 m/s/s)(3.78 s)^{2}

v_{i}= 12.781 m/s = 12.8 m/s

For the velocity of impact with the ground, let's use

v_{f}= v_{i}+ a(Dt)

v_{f}= (12.781 m/s) + (-9.8 m/s/s)(3.78 s) = -24.2627 m/s = -24.3 m/s

For how high the ball goes we have to start over:

Dx = ?? (Don't know anymore)

v_{i}= 12.781 m/s

v_{f}= 0 m/s (It is at the top)

a = -9.8 m/s/s (gravity)

Dt = ??? (don't really care)

I would use:

v_{f}^{2}= v_{i}^{2}+ 2a(Dx)

(0)^{2}= (12.781 m/s)^{2}+ 2(-9.8 m/s/s)(Dx)

Dx = 8.335 m = 8.33 m above the building.

For its position at 2.12 seconds, use:

Dx = v_{i}(Dt) +^{1}/_{2}a(Dt)^{2}

Dx = (12.781 m/s)(2.12 s) +^{1}/_{2}(-9.8 m/s/s)(2.12 s)^{2}

Dx = 5.074 = 5.07 m (above the building still)

And its velocity is:

v_{f}= v_{i}+ a(Dt)

v_{f}= (12.781 m/s) + (-9.8 m/s/s)(2.12 s) = -7.995 = -8.00 m/s (going down)

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**4. Catona Hotinruff is ascending in a helicopter at a
rate of 13.3 m/s. At an elevation of 47.0 m, she drops a bagel out the window of
the 'copter. What time does the bagel take to reach the ground? What is its
velocity of impact? How high is the helicopter when the bagel hits the ground?**

For the bagel from when it leaves the 'copter, to when it reaches the ground we know:

Dx = -47.0 m (Down is negative)

v_{i}= +13.3 m/s

v_{f}= ??

a = -9.8 m/s/s (gravity)

Dt = ??

Now, I'm lazy, and I don't want to use the formula:

Dx = v_{i}(Dt) +^{1}/_{2}a(Dt)^{2}

as that would be very hard to solve for time, so let's go after the final velocity:

v_{f}^{2}= v_{i}^{2}+ 2a(Dx)

|v_{f}| = Ö{v_{i}^{2}+ 2a(Dx)} = Ö{(13.3 m/s)^{2}+ 2(-9.8 m/s/s)(-47.0)} = 33.14 m/s, so the answer could be +33.14 m/s, or -33.14 m/s, but we know it is going down when it hits the ground, so the correct choice is -33.14 m/s.

Now the time is easy:

v_{f}= v_{i}+ a(Dt)

-33.14 m/s = (13.3 m/s) + (-9.8 m/s/s)(Dt)

Dt = 4.739 = 4.74 s

Assuming the 'copter keeps ascending at 13.3 m/s, it will be

v_{avg}= Dx/Dt

(13.3 m/s) = Dx/(4.739 s)

Dx = 63.02 m higher than it was, which was 47.0 m above the ground, so the 'copter is 47.0 m + 63.02 m = 110 m above the ground

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**5. Colin Host is driving his Ferrari 345 m behind my
Tercel. I am going 28.2 m/s, but the Colin overtakes me in 16.43 seconds. How
fast is the Ferrari going? How far does the Ferrari travel before it overtakes
me? **

Colin is approaching me at

v_{avg}= Dx/Dt = (345 m)/(16.43 s) = 20.998 m/s = 21.0 m/s, which means he is going that much faster than me, so his velocity must be 28.2 m/s + 21.0 m/s = 49.2 m/s, which is way too fast. He must have never done the No reason to speed lab.

His car then must go:

v_{avg}= Dx/Dt

(49.2 m/s) = Dx/(16.43 s)

Dx = 808.326 m = 808 m

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**6. Bob White is 12.5 miles from home. If he drives the
first 3.0 miles home at 35 mph, how fast does he need to drive the rest to
average 55 mph? **

If he averages 55 MPH, then the trip home must take

v_{avg}= Dx/Dt

(55 mi/hr) = (12.5 mi)/Dt

Dt = 0.227273 hr

If he has driven already 3.0 mi at 35 MPH, he has used up:

v_{avg}= Dx/Dt

(35 mi/hr) = (3.0 mi)/Dt

Dt = 0.08571 hr

Which means he has

0.227273 hr - 0.08571 hr = 0.14156 hr left to go the remaining 9.5 mi if he is going to average 55 MPH (By completing the trip in .227273 hr)

So his speed must be:

v_{avg}= Dx/Dt

v_{avg}= (9.5 mi)/(0.14156 hr) = 67.1 MPH

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**7. Justin Case fires an air rocket upward at 34.5 m/s
from ground level. Unfortunately (for him) the rocket lands on the top of a 32.0
m tall light tower on its way down. What time is the rocket in the air? How high
does the rocket go? With what velocity does it strike the tower? **

For the rocket leaves the ground, to when it strikes the light tower, we know:

Dx = 32.0 m

v_{i}= +34.5 m/s

v_{f}= ??

a = -9.8 m/s/s (gravity)

Dt = ??

Let's find the velocity it strikes the tower first: (it's easier)

v_{f}^{2}= v_{i}^{2}+ 2a(Dx)

| v_{f}| = Ö{ v_{i}^{2}+ 2a(Dx)} = Ö{(34.5 m/s)^{2}+ 2(-9.8 m/s/s)(32.0)} = 23.73 m/s, so the answer could be +23.73 m/s, or -23.73 m/s, but we know the correct answer is -23.73 m/s as it is going down when it hits the tower.

Now we can find the time of impact:

v_{f}= v_{i}+ a(Dt)

-23.73 m/s = (34.5 m/s) + (-9.8 m/s/s)(Dt)

Dt = 5.942 s = 5.94 s

For how high the rocket goes, we start over:

Dx = ??

v_{i}= +34.5 m/s

v_{f}= 0 m/s (it is at the top)

a = -9.8 m/s/s (gravity)

Dt = ?? (don't care)

Use:

v_{f}^{2}= v_{i}^{2}+ 2a(Dx)

(0)^{2}= (34.5 m/s)^{2}+ 2(-9.8 m/s/s)(Dx)

Dx = 60.7 m

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**8. Molly Fayad pops a softball up that ends up being
caught by the catcher at the same elevation as she hit it. A spectator, using a
sextant, and a range finder, determines that the ball went 63.0 m into the air
at its highest point. For what total time was the ball in the air? At what times
is the ball at an elevation of 42.0 m? **

Let's look at the trip from the bat to the top

Dx = 63.0 m (given)

v_{i}= ??

v_{f}= 0 m/s (it is at the top)

a = -9.8 m/s/s (gravity)

Dt = ??

Use this to find the initial velocity:

v_{f}^{2}= v_{i}^{2}+ 2a(Dx)

(0)^{2}= v_{i}^{2}+ 2(-9.8 m/s/s)(63.0 m)

v_{i}= 35.14 m/s or -35.14 m/s, but we know it was going up to start with, so the correct answer is +35.14 m/s.

Hmmmm - it doesn't ask for initial velocity, but I bet it is useful.

Let's find the time to the top:

v_{f}= v_{i}+ a(Dt)

0 = (35.14 m/s) + (-9.8 m/s/s)(Dt)

Dt = 3.5857 s, and the total time is twice this or 7.17 s

At elevation 42.0 m we need to start over:

Dx = 42.0 m (given)

v_{i}= 35.14 m/s

v_{f}= ??

a = -9.8 m/s/s (gravity)

Dt = ??

Let's solve for the final velocity, as solving for time is quadratic:

v_{f}^{2}= v_{i}^{2}+ 2a(Dx)

|v_{f}| = Ö{ v_{i}^{2}+ 2a(Dx)} = Ö{(35.14 m/s)^{2}+ 2(-9.8 m/s/s)(42.0)} = 20.288 m/s, so the answer could be +20.288 m/s, or -20.288 m/s, and in this case, both are true, the first on the way up, and the second on the way down. Now we just need to find the times. Let's do 42.0 m on the way up first:

Dx = 42.0 m (given)

v_{i}= 35.14 m/s

v_{f}= +20.288 m/s

a = -9.8 m/s/s (gravity)

Dt = ??

Find time with:

v_{f}= v_{i}+ a(Dt)

+20.288 m/s = 35.14 m/s+ (-9.8 m/s/s)(Dt)

Dt = 1.52 s for on the way up,

On the way down

Dx = 42.0 m (given)

v_{i}= 35.14 m/s

v_{f}= 120.288 m/s (going down)

a = -9.8 m/s/s (gravity)

Dt = ??

Find time with:

v_{f}= v_{i}+ a(Dt)

-20.288 m/s = 35.14 m/s+ (-9.8 m/s/s)(Dt)

Dt = 5.66 s for on the way down

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**9. Austin Tascious sends a bowling ball down a 16.5 m
long lane, and we hear the sound of impact 2.53 seconds later. If the speed of
sound is 343 m/s, what speed did the bowling ball move? **

We know that the 2.53 seconds is divided into two parts, first, the time the ball takes to reach the pins, and second, the time the sound, traveling at 343 m/s takes to get back.

First, let's find the time the sound takes:

v_{avg}= Dx/Dt

343 m/s = (16.5 m)/Dt

Dt = 0.0481 s

Now we know that the ball took

2.53 s - 0.0481 s = 2.482 s

Finally, we can use the formula for speed to calculate the speed of the ball:

v_{avg}= Dx/Dt

v_{avg}= (16.5 m)/(2.482 s) = 6.65 m/s

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**10. Ali Katz is taking a dripping paint bucket back to
the store, so he holds the paint bucket out the car's window, letting the paint
drip onto the ground. The bucket is dripping at some unknown rate, and his car
accelerates at some unknown rate, starting from rest. If the first drop drips
when his car's velocity is zero, the second drop is 2.3 m down the road, how far
is the sixth drop down the road? **

The problem is that we don't know the acceleration of the car or the time between drips. Either the problem is unsolvable, or, it doesn't matter, as long as the drip rate and acceleration are constant.

Here's what we know, if the time between drops is t, then the second drop happened 1t after the first drop, and the sixth drop occurred 5t seconds after the first drop.

Let's plug it into:

Dx = v_{i}(Dt) +^{1}/_{2}a(Dt)^{2}

For the second drop:

2.3 m = (0)(1t) +^{1}/_{2}a(1t)^{2 }=^{1}/_{2}a(t)^{2}

For the sixth drop:

Dx = (0)(5t) +^{1}/_{2}a(5t)^{2}=^{1}/_{2}a(5t)^{2 }= (25)^{1}/_{2}a(t)^{2}

So now we have two equations and two unknowns:

2.3 m =^{1}/_{2}a(t)^{2}

Dx = (25)^{1}/_{2}a(t)^{2}

Hmm, doesn't it follow that

Dx = (25)(2.3) = 57.5 m?

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