Friday

November 27, 2015
Posted by **Anonymous** on Saturday, March 15, 2008 at 3:21pm.

a) the time taken to move to A

b) the distance between A and the building

c) the horizontal and vertical components of velocity at B.

d) the time required to move from A to B

e) the object's speed at B.

So I started out by finding the initial velocity components and the horizontal and vertical components are both 15m/s because 21.21sin45 = 15 and 21.21cos45 = 15.

At A the parallel component remains the same so it will be 15m/s. Not sure how to find the rest of the info as I only know that it's 20m below the level of the top of building...

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**Damon**, Saturday, March 15, 2008 at 3:34pmRemember if a = -10 and vertical speed is v then that

v = Vo - 10 t

h = ho + Vo t - 5 t^2

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**Anonymous**, Saturday, March 15, 2008 at 3:55pmI don't understand. They don't give us original height of the building.

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**Damon**, Saturday, March 15, 2008 at 3:58pmSo I called the original height zero and worked from there. Therefore A is at NEGATIVE 20 meters

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**Damon**, Saturday, March 15, 2008 at 4:03pmBy the way, in problems with potential energy like height, you can start at any value at all. Only changes matter. I could have called the roof 1000 meters high, the A would have been at 980 meters .

All that matters is that A is 20 meters below the starting point.

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**Damon**, Saturday, March 15, 2008 at 4:21pmIf I had called the roof at h = 1000 meters, then my equation for h at A would be:

980 = 1000 + 15 t - 5 t^2

or once again as before

-20 = 15 t - 5 t^2

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**Damon**, Saturday, March 15, 2008 at 3:37pmcall ho = 0 at the roof where we started

call Vo = 15 which is the initial vertical speed you calculated

the

Until it hits the ground

v = 15 - 10 t

h = 15 t - 5 t^2

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**Damon**, Saturday, March 15, 2008 at 3:40pmpoint A is at h = -20

so

h = -20 = 15 t - 5 t^2

5 t^2 - 15 t - 20 = 0

t^2 - 3 t - 4 = 0

(t-4)(t+1) = 0

t = 4 or t = -1 we do not do negative time so

t = 4 seconds at A

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**Damon**, Saturday, March 15, 2008 at 3:42pmHow far did it go horizontal in 4 seconds from the building

horizontal speed you told me was 15 m/s until we crash so

horizontal distance from A to building = 15*4 = 60 meters

Need I continue or do you get it now?

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**Anonymous**, Saturday, March 15, 2008 at 4:05pmI think i get it now. At B it says it moves at angle of 18.43degrees to the vertical. How would you calculate components when it's to the vertical? Would it be

to calculate the hyp = 15/(sin15.83) = 55m/s

55x(cos15.83)

= 53m/s

So the vertical component is 53m/s...that doesn't seem right.

d) how would you calculate the time required from A to B?

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**Damon**, Saturday, March 15, 2008 at 4:15pmAt B as at A, the horizontal speed is 15 m/s still

I want vertical speed, v

the tangent of 18.43 = opposite/adjacent = 15/|v| ( using absolute value because in my triangle I know v is negative, down)

.333 = 15/|v|

|v| = 45

so

v = -45 or 45 meters/second down

Now, remember

v = 15 - 10 t

-45 = 15 - 10 t

t = 6 seconds after start so 2 seconds after A

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**Anonymous**, Saturday, March 15, 2008 at 4:40pmThanks so much for all the help!

For e) do I just use the pythagorean theorem to find the speed?

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**Damon**, Saturday, March 15, 2008 at 4:52pmsure, 15 horizontal, -45 vertical so

sqrt(225 + 2025)