Posted by Anonymous on Saturday, March 15, 2008 at 3:21pm.
Remember if a = -10 and vertical speed is v then that
v = Vo - 10 t
h = ho + Vo t - 5 t^2
I don't understand. They don't give us original height of the building.
So I called the original height zero and worked from there. Therefore A is at NEGATIVE 20 meters
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.
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
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
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
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?
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?
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
Thanks so much for all the help!
For e) do I just use the pythagorean theorem to find the speed?
sure, 15 horizontal, -45 vertical so
sqrt(225 + 2025)
Related Questions
Physics - A ball is thrown at an angle of 37.1degrees above the horizontal from ...
Physics - An objectis thrown upward from the edge of a tall building with a ...
physics - If an object is thrown in an upward direction from the top of a ...
physics - A ball is thrown from the top of a building upward at an angle of 66 &...
Physics - An object is thrown upward with a speed of 21.5 m/s. How high above ...
Physics - A brick is thrown upward from the top of a building at an angle of 27....
Physics - A rock is thrown vertically with a velocity of 21 m/s from the edge of...
physics - A marble is thrown horizontally with a speed of 10.5 m/s from the top ...
Physics - An object is projected at an angle of 53.1degrees above the horizontal...
physics - Three identical balls are thrown from the top of a building, all with ...
For Further Reading
