# Physics

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An object is thrown from the top edge of a building at a speed of 21.21 m/s at an angle of 45degrees above the horizontal. Some time later it passes point A which is 20m below the level of the top of the building, and continues to strike ground at point B. It reaches B moving at an angle of 18.43degrees to the vertical. Determine:

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...

• Physics - ,

Remember if a = -10 and vertical speed is v then that
v = Vo - 10 t
h = ho + Vo t - 5 t^2

• Physics - ,

I don't understand. They don't give us original height of the building.

• Physics - ,

So I called the original height zero and worked from there. Therefore A is at NEGATIVE 20 meters

• Physics - ,

By 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.

• Physics - ,

If 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

• Physics - ,

call 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

• Physics - ,

point 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

• Physics - ,

How 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?

• Physics - ,

I 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?

• Physics - ,

At 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

• Physics - ,

Thanks so much for all the help!

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

• Physics - ,

sure, 15 horizontal, -45 vertical so
sqrt(225 + 2025)