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March 30, 2017

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From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 9.40 m/s and angle of 23.0° below the horizontal. It strikes the ground 6.00 s later.

Initial velocity components:
vx = 8.65 m/s
vy = -3.67 m/s

(c) Find the equations for the x- and y- components of the position as functions of time.
x = 8.65t
y = ?

(d) How far horizontally from the base of the building does the ball strike the ground?
51.9 m

(e) Find the height from which the ball was thrown.

(f) How long does it take the ball to reach a point 10.0 m below the level of launching?

I need help on (c), (e), and (f).

  • Physics - ,

    Vo = 9.4m/s[23o]
    Xo = 9.4*cos23 = 8.65 m/s.
    Yo = 9.4*sin23 = 3.67 m/s.

    c. X = Xo * t = 8.65t
    Y = Yo*t + 0.5g*t^2=3.67t+4.9t^2

    e. h = 3.67t + 4.9t^2
    h = Y = 3.67*6 + 4.9*6^2 = 198.4 m.
    Above gnd.

    f. 3.67t + 4.9t^2 = 10 m.
    4.9t^2 + 3,67t _- 10 = 0
    t = 1.10 s. (Use Quad. Formula).

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