mmiller39
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I will start with the one that I have worked out already. I am submitting the answers to an online testing site and the answer that I have submitted is wrong but I believe that my math and selection of proper equations is correct.
THE QUESTION:
A pellet gun is fired straight downward from the edge of a cliff that is 11.4 m above the ground. The pellet strikes the ground with a speed of 27.5 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?
MY WORK:
First we have to find the initial velocity (which will be the same in both situations).
I used: V^2 = Vo^2 + 2(a)(y)
Which is 27.5^2 = Vo^2 + 2(-9.80)(11.4)
Vo = 31.3 m/s
Then use the equation with V = 0 as it is at the peak of the shot.
y = V^2 - Vo^2/(2(-9.80))
-31.30^2/-19.6 = 49.984 m <----------- WRONG ANSWER, anything you see wrong?
SECOND QUESTION:
You live in a building on the left, and a friend lives in a building on the right. The two of you are having a discussion about the heights of the buildings, and your friend claims that his building is half again as tall as yours. To resolve the issue you climb to the roof of your building and estimate that your line of sight to the top edge of the other building makes an angle of 21° above the horizontal, while your line of sight to the base of the other building makes an angle of 52° below the horizontal.
http://edugen.wiley.com/edugen/courses/crs1507/art/qb/qu/c01/ch01p_62.gif
(a) Determine the ratio of the height of the taller building to the height of the shorter building.
MY WORK:
I have no Idea where to begin to find the solution to this, I have tried 21/52, that does not work. Any suggestions?
Thanks so much guys(and girls)
THE QUESTION:
A pellet gun is fired straight downward from the edge of a cliff that is 11.4 m above the ground. The pellet strikes the ground with a speed of 27.5 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?
MY WORK:
First we have to find the initial velocity (which will be the same in both situations).
I used: V^2 = Vo^2 + 2(a)(y)
Which is 27.5^2 = Vo^2 + 2(-9.80)(11.4)
Vo = 31.3 m/s
Then use the equation with V = 0 as it is at the peak of the shot.
y = V^2 - Vo^2/(2(-9.80))
-31.30^2/-19.6 = 49.984 m <----------- WRONG ANSWER, anything you see wrong?
SECOND QUESTION:
You live in a building on the left, and a friend lives in a building on the right. The two of you are having a discussion about the heights of the buildings, and your friend claims that his building is half again as tall as yours. To resolve the issue you climb to the roof of your building and estimate that your line of sight to the top edge of the other building makes an angle of 21° above the horizontal, while your line of sight to the base of the other building makes an angle of 52° below the horizontal.
http://edugen.wiley.com/edugen/courses/crs1507/art/qb/qu/c01/ch01p_62.gif
(a) Determine the ratio of the height of the taller building to the height of the shorter building.
MY WORK:
I have no Idea where to begin to find the solution to this, I have tried 21/52, that does not work. Any suggestions?
Thanks so much guys(and girls)