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March 28, 2017

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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)

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