How Far II : The Sequel
by Dustin Glazier, January 1998

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1.

A baseball leaves the bat with an upward velocity of 54 m/s.
How high does it go? What time does it take to reach the top?
What total time will it be in the air? (149 m, 5.5 s, 11s) Here is what you start with:

Well let's find an equation that solves for X Of the many that do fit, 2aX = V2 - Vi2 is best because we know all the variables exept for X. Plug in those cute numbers: 2(-9.8 m/s2)X = (0)2 - (54 m/s)2 X = -2916 m/s ------------- -19.6 m/s2 X = 148.77 which is only 149m with sig figs
We now move on to the time it takes for it to reach the top: I think that X = Vit + 1/2at2 looks appealing so let's try it.

We then plug in the numbers:
149 = 0t + 1/2(-9.8 m/s2)t2 Finishing it through: 149 = -4.9t2 Leaving us with t2 = 30.40816327 take the square root of that and it gives you t = 5.5 seconds And to find the total time in the air we simply double the time to get to the top> Because the acceleration is the same for both going up and down, so is the time to reach
Therefore the time for the whole flight is t = 11 seconds 
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2.

 A person jumps off of a cliff and hits the water below moving
with a velocity of -24 m/s. What time were they in the air?
How high is the cliff? (2.4 s, 29 m) Here is what you start with: Let's plug this in to the formula that we know all the variables for, except for t V = Vi + at, seems the most appropriate So, -24 m/s = (0) + (-9.8m/s2)t Therefore we divide -24 by 9.8 and it gives us t = 2.448979592 seconds With this we can now find the height of the cliff With X = Vat Va = -12 m/s because Va = (Vi + V)/2 and X = (-12 m/s)(2.4 s) Therefore X = 29m but we know that the negative sign denotes that the jumper went down.
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3.

 Cliff divers in South America jump from 300 foot cliffs into
the water. (1 m = 3.281 f) What time does it take them to hit the water,
and how fast are they going
when they do hit the water? (4.3 s, 42 m/s) Here is what you start with: We find a formula X = Vit + 1/2at2 seems like the best one So we put in the numbers: 91.44 = (0)t + 1/2(9.8 m/s2)t2 91.44 m = 4.9t2 18.661 = t2 with the square root we find t = 4.3 seconds Now moving on to impact velocity Using the formula Vf = Vi + at Plug it in, Plug it in aaannnnddd Vf = 42.238 or 42 m/s 
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4.

 Red Elk leaves the 10.0 m diving board with an
upward velocity and hits the water 1.9 seconds later. What was his initial upward velocity?
To what height above the diving board did he rise before going down?
With what speed did he hit the water? (4.0 m/s, .84 m, -14.6 m/s) Here is what you start with: We first have to find an appropriate formula.
Since were solving for Vi lets find one that solves for it.
X = Vit + 1/2at2 is quite appetizing lets see if it solves only for Vi
10m = Vi(1.9s)+ 1/2(-9.8 m/s/s)(1.9)2 those arithmetic skills yield Vi = 4.0468 m/s which with sig figs is 4.0 m/s Now, since we know that at the highest point his V = 0, we can use that to solve for X With 2aX = V2 - Vi2 and the numbers make it look like this: 2(-9.8 m/s2)X = 02 - (4.04 m/s)2 again we use those algebra skills and get X = .84 m But we're not done yet, we still must find Red Elk's velocity on impact We'll use the equation Vf2 = Vi2 + 2aX Since the Vi is from the highest point then the Vi = 0 Then we fill in the rest of the equation Vf2 = 0 + 2(-9.8 m/s2)(10.84)
It's important to remember that the X = 10.84 because that ís the highest point of the arc
Because the original X = 10 m and the distance higher is .84 m Working it through and remembering that it ís negative because he's going down:
Vf = -14.576 or -14.6 m/s Cool, it ís done!! 
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5.

 A car will skid to a halt at a rate of -9.4 m/s/s.
If you measure skid marks that are 34 m long, with what speed
was the car going that made them? (25 m/s) Here is what you start with: This is a simple acceleration equation: Since we donít have any time given the obvious equation is Vf2 = Vi2 + 2aX Weíre trying to find the original velocity an the plugged in numbers: 02 = Vi2 + 2(-9.4 m/s/s)(34 m) That handy dandy math and Vi = 25.2824 or significantly 25 m/s! 
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6.

 A train can speed up at .15 m/s/s. In what minimum
distance can it attain a speed of 25 m/s starting from rest?
(2083 m) Here is what you start with: Hey, Iíve got an idea, letís find a formula to use We seem to use Vf2 = Vi2 + 2aX alot, and it seems best.
Linda, something from the meat cabinet. Or perhaps the number cabinet. 252 = 02 + 2(.15 m/s/s)X Break it down homeboy, and you get X = 2083.3333333333333333333 or 2083 m 
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7.

 A drag racer can reach a speed of 53 m/s over
a distance of 120 m. What is its acceleration? Over what distance
can it reach a speed of 85 m/s? (11.7 m/s/s, 309 m) Here is what you start with: I tell ya, I just canít get enough of that Vf2 = Vi2 + 2aX With some semi-redundant plugging in of numbers we have (53 m/s)2 = 02 + 2a(120 m) Hey, uhhhh, letís try solving it through!! Lo and behold, a = 11.7 m/s/s What, thereís more, letís see what formula might be a good idea. Since we donít know t weíll use the formula: r2 = 1/(mv2/2ke2) + (1/r1) Ha, Ha, Ha, thatís not until electric energy. Instead we use good oleí Vf2 = Vi2 + 2aX Those numbers plugged in, remembering that the Vf = 85 m/s gives us (85 m/s)2 = 02 + 2(11.7 m/s/s)X And if we can still remember how to do algebra, than we find X = 308.76068 or 309 m 
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8.

 A jetliner must reach a speed of 80 m/s to take off,
and can accelerate at 4.7 m/s/s. What is the minimum length of runway? (681 m) Here is what you start with: Since we donít know the amount of time than we must again use Vf2 = Vi2 + 2aX Those numbers go in and make the lovely little equation (80 m/s)2 = 02 + 2(4.7 m/s/s)X Work it on down and you are left with X = 680.8510638 or 681 m 
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9.

 Theoretically, what would be the velocity of a steel marble dropped from an airplane
1000 m above the ground just as it hits the ground? (-140 m/s) Here is what you start with: Use the formula Vf2 = Vi2 + 2aX This formula seems best for most of these problems because they donít give you time Anyway, plug the stuff in to have Vf2 = 02 + 2(9.8 m/s/s)(1000 m) bust a move and get Vf = 140 or -140 m/s (because itís going down!) 
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10.

 A rifle bullet leaves the muzzle of a .75 m long barrel going 450 m/s.
What is the acceleration of the bullet while it is in the barrel? (135,000 m/s/s) Here is what you start with: Use Vf2 = Vi2 + 2aX We then have (450 m/s)2 = (0 m/s)2 + 2a(.75 m) 202500 = 1.5a a = 135000 m/s/s 
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Hey this page was not easy, but I feel privileged to have been able to help you grasp
Physics. And I really needed the extra credit! Happy Solving!!