Expert: James J. Kovalcin - 11/17/2004

Question
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Followup To
Question -
If a football is thrown at 5.00m/s off a 30.0m high cliff at 30.0 degrees upward. How long does it take to hit the ground? How high does the ball go from the base of the cliff? How far from the base of the cliff does it go? How fast will the ball be going when it is even with the cliff on its decent? What is the overall velocity of the ball just before it hits the ground?


Answer -
All problems of this type are solved in the same way!
Set up two data table listing the 6 possible variables, one for the horizontal and a second for the vertical.
In this case for the vertical: Do=30m, Df=?, Vo=5*sin30=2.5m/s, Vf=?, a=-9.8m/s^2 and t=?.
For the horizontal: Doh=0m, Dfh=?, Voh=5*cos30=4.3m/s, Vfh=4.3m/s, ah=0m/s^2 and t=?.
Since you always have two equations available in any uniform acceleration problem:
Df=1/2*a*t^2+Vo*t+Do and Vf=a*t+Vo
All you need to do is to determine four of the six variables [since you are allowed two unknowns with two equations].
Let's go to the first question.
1. How long does it take to hit the ground?
The only new piece of information that this clue supplies is that the final vertical position of the ball will be the ground. Therefore, Df=0m. You now MUST use this new piece of information. Since the only equation containing this variable is Df=1/2*a*t^2+Vo*t+Do, make Df=0 and solve for the time using the quadratic formula.
2. How high does the ball go from the base of the cliff?
This question offers an entirely differrent clue! At the highest point reached by the ball, the only special piece of information that you have is that the final vertical velocity will be Vf=0m/s! And again, since this is the only special piece of information you must use it. The only equation containing Vf is Vf=a*t+Vo. Make Vf zero and solve for the time to the highest point. Then use this time in the displacement equation to determine the height of the ball above the ground, Df=1/2*a*t^2+Vo*t+Do.
3. To calculate the horizontal range just multiply the time to the ground from question #1 above by the constant horizontal velocity of 4.3m/s. Df=v*t [the long equation where a and Do are both zero.
4. How fast will the ball be going when it is even with the cliff on its descent?
There are two different ways to handle this. The first way is to make Df equal to Do and use Df=1/2*a*t^2+Vo*t+Do to determine the time [easy because this does not require the quadratic formula]. The use this time to calculate the final vertical velocity. Finally, combine this vectorally with the constant horizontal velocity to determine the speed using the Pythagorean Theorem. After doing all of this there is an easier way! SInce the acceleration of gravity is the same both on the way up and on the way down, from a symmetry point of view Vf must equal Vo or 30m/s!
5. What is the overall velocity of the ball just before it hits the ground?
To answer this question take the time to the ground that you calculated in question #1 and use it to calculate the final vertical velocity of the ball using Vf=a*t+Vo then add this answer vectorally to the same constant horizontal velocity. Don' forget that velocity is a vector and you need to find both the magnitude of the velocity [from the Pythagorean Theorem] and the direction [by using the inverse tangent].


Thank you for all of your help! very good directions! I was wondering could you check my answeres to see if I did everything correctly and kept the correct number of significant digits.

1. 0=-4.9t^2+2.5t = 30.0 after using the quadratic formula I got 5.20s
2. vf=at+vo after rearranging t=vf-vo/a t=0-2.5/-9.8)=.255s
Df=1/2(-9.8)(.255)+(-2.50)(.255)+30=29.0m
3.Df=Vt
(5.20)(4.33)=22.5m
4. 30.0m/s because of the accelaration of gravity
5.vf=at+vo vf=(9.8)(5.20)+2.50= 53.3m/s vertically
vf=at+vx vf=(9.8)(5.20)+4.33= 55.3m/s horizontally
53.5^2=55.3^2= Total^2= 76.9m/s
inv tan is 44.1 degrees


Answer
1. I get a time of 3.76s, check your quadratic formula.
2. You forgot to square the time in 1/2*a*t^2. I get 30.3m [has to be higher than the cliff!]
3. Use a time of 3.76s from #1.
4. Same as the initial velocity due to symmetry.
5. THe gravitational acceleration should be negative! -9.8, final vertical velocity -34m/s.
The horizontal velocity remains a constant 4.3m/s since there is no horizontal acceleration.
To find the final velocity use the pythagorean thorem.