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**Falling object - by Kevin Bailey & Chicken Breast,
2002**

**1. A robot probe drops a camera off the rim of a 239 m
high cliff on Mars, where the free-fall acceleration is -3.7 m/s ^{2}.**

** a. Find the time required for the camera to reach
the ground.**

** b. Find the velocity with which it hits the
ground.**

Ahhh, those crazy robots from the future. Always hijacking NASA's rockets, traveling to Mars, and dropping government property off cliffs at a slower acceleration than Earth.

Any who, if you wish to find the time, just talk to pooh, he'll tell you. Dx = v

_{i}(Dt) +^{1}/_{2}a(Dt)^{2}Dx = 239 m

v

_{i}= 0 m/s

Dt = ???

v_{f} = ???

a = -3.7 m/s^{2}

So, (239 m) = (0 m/s)(Dt) + ^{1}/_{2}(-3.7
m/s^{2})(Dt)^{2} and (Dt)
= **11.4 s **oH yEs LiFe Is SwEeT

Furthermore, if you wish to find the final velocity just use v_{f} = v_{i}
+ a(Dt)

Dx = 239 m

v

_{i}= 0 m/s

Dt = 11.4 s

v_{f} = ???

a = -3.7 m/s^{2}

So, (v_{f}) = (0 m/s)_{ + }(-3.7 m/s^{2})(11.4
s) = **-42.18 m/s or -42.1 m/s**

**2. A flowerpot falls from a windowsill 25.0 m above the
sidewalk.**

** a. How fast is the flowerpot moving when it
strikes the ground?**

** b. How much time does a passerby on the sidewalk
below have to move out of the way before the flowerpot hits the ground?**

Boy, I
sure hope that the poor passerby doesn't get hit! ** **

Yeah, well I'm just gonna do this one backwards. Time first. Dx = v

_{i}(Dt) +^{1}/_{2}a(Dt)^{2}Dx = 25 m

v

_{i}= 0 m/s

Dt = ???

v_{f} = ???

a = -9.8 m/s^{2}

So, (25 m) = (0 m/s)(Dt) + ^{1}/_{2}(-9.8
m/s^{2})(Dt)^{2} and (Dt)
= **2.26 s **Chicken Breast

Furthermore, if you wish to find the final velocity just use v_{f} = v_{i}
+ a(Dt)

Dx = 25 m

v

_{i}= 0 m/s

Dt = 2.26 s

v_{f} = ???

a = -9.8 m/s^{2}

So, (v_{f}) = (0 m/s)_{ + }(-9.8 m/s^{2})(2.26
s) = **-22.1 m/s**

**3.A tennis ball is thrown vertically upward with an
initial velocity of +0.80 m.**

** a. What will the ball's speed be when it returns
to its starting point?**

** b. How long will the ball take to reach its
starting point?**

So, what we have here is really 2 equations. 1 on the way up, and 1 on the way down. BUT, because TIME IS ON BOTH SIDES...that's right, it takes just as long to go up as down, we will just solve for up and double. Use v

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

Dx = ???

v

_{i}= 8.0 m/sDt = ???

v

_{f}= 0 m/sa = -9.8 m/s

^{2}So, (0 m/s) = (8.0 m/s) + (-9.8 m/s

^{2)}(Dt) so Dt =1.63 sDon't forget to double it!

OH yeah, and part A. Well, I think that

-8.0 m/s, the same as the equal but opposite velocity as the starting one I think is pretty much self explanitory.

**4. Stephanie hits a volleyball from a height of 0.80 m
and gives it an initial velocity of +7.5 m/s straight up.**

** a. How high will the volleyball go?**

** b. How long will it take the ball to reach its
maximum height? (Hint: At maximum height, v = 0 m/s)**

Humm...this is a tricky one. In order to find the displacement you need to know this formula. v

_{f}^{2}= v_{i}^{2}+ 2a(Dx), where:Dx = ???

v

_{i}= 7.5 m/sDt = ???

v

_{f}= 0 m/sa = -9.8 m/s

^{2}So just plug in the numbers and let that magical math machine do the rest!

(0 m/s)

^{2}= (7.5 m/s)^{2}+ 2(-9.8 m/s^{2})(Dx) so Dx =2.87 mOk, and for the next part you will take heed to the hint, and use the formula v

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

Dx = 2.87 m

v

_{i}= 7.5 m/sDt = ???

v

_{f}= 0 m/sa = -9.8 m/s

^{2}So again, just plug in the numbers and let that magical math machine do the rest!

(0 m/s) = (7.5 m/s) + (-9.8 m/s

^{2})(Dt) so Dt =.765 s

**5. Maria throws an apple vertically upward from a height
of 1.3 m with an initial velocity of +2.4 m/s.**

** a. Will the apple reach a friend in a treehouse 5.3 m
above the ground?**

** b. If the apple is not caught, how long will the apple
be in the air before it hits the ground? **

Who do you suppose built that treehouse?

This question is basically a pretty clever way to ask for the height. Don't forget to subtract Maria's 1.3 m height! Use the following formula. v

_{f}^{2}= v_{i}^{2}+ 2a(Dx), where:Dx = ???

v

_{i}= 2.4 m/sDt = ???

v

_{f}= 0 m/sa = -9.8 m/s

^{2}Uh oh, I hear that familiar chomping sound of the Math Robot coming again! Don't worry, just plug and chicken breast it.

(0 m/s)

^{2}= (2.4 m/s)^{2}+ 2(-9.8 m/s^{2})(??? m) so height =.29 m + 1.3 m = 1.6 m NO, WON'T REACHWell remember for the second part that time is on both sides, it takes just as long to go up as down, we will just solve for up and double. Use v

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

Dx = ???

v

_{i}=2.4 m/sDt = ???

v

_{f}= 0 m/sa = -9.8 m/s

^{2}So, (0 m/s) = (2.4 m/s) + (-9.8 m/s

^{2)}(Dt) so Dt =.4897959 sDon't forget to double it! But we're not done. The apple started at 1.3 m, so we still have that much farther to fall.Dx = 1.3 m

v

_{i}=-2.4 m/sDt = ???

v

_{f}= ??? m/sa = -9.8 m/s

^{2}So using Dx = v

_{i}(Dt) +^{1}/_{2}a(Dt)^{2}then (1.3 m) = (2.4 m/s)(Dt) +^{1}/_{2}(-9.8)(Dt)^{2}Dt =

.82 s

**6. Calculate the displacement of the volleyball in sample
Problem 2F when the volleyball's final velocity is 1.1 m/s upward. **

<Insert Answer Here>

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