Physics



From the top of a cliff with 80 m height a stone is thrown vertically and finally hit the surface of ocean.
Initial velocity of stone is 40 m/s.

Find:

(a) time to reach to max height.
(b) how high goes from surface of ocean?
(c) what is velocity of stone 2 second after thrown?
(d) what is velocity of stone at 20 meters above ocean?
(e) total traveled time?
(f) at what times velocity is 10 m/s?
(g) what is velocity of ball 40 m above ocean?
(h) what is velocity of ball after 6 second from thrown?
(i) at what height from the ocean surface velocity is -50 m/s?
(j) what is velocity of stone just before hit the ocean surface?


I calculated that:
(a) tmax = 4.08 s
(b) ytotal = 161.6 m
(c) vy = 20.4 m/s
(d) vy = 20.4 m/s
(e) ttotal = 9.83 sec
(f) t1 = 3.06 sec ; t2 = 5.1 sec


How would you exactly find g~j? :/
I suppose the variables confuses me a bit.

asked by Lilian
  1. a. Tr = -Vo/g = -40/9.8 = 4.08 s. = Rise
    time.

    b. h = ho + (V^2-Vo^2)/2g
    h = 80 + (0-(40^2))/-19.6 = 161.6 m.
    Above gnd.

    c. V = Vo + g*t = 40 - 9.8*2 = 20.4 m/s.

    d. V*2 = Vo^2 + 2g*(h-20)
    V^2 = 0 + 19.6*(161.6-20) = 2775.36
    V = 52.7 m/s.

    e. h = 4.9t^2 = 161.6 m.
    4.9*t^2 = 161.6
    t^2 = 32.979
    Tf = 5.74 s. = Fall time.

    Tr+Tf = 4.08 + 5.74 = 9.82 s. = Total
    time traveled.

    f. V = Vo+g*t = 40 - 9.8*t = 10 m/s.
    -9.8t = 10-40 = -30
    t = 3.06 s.

    g. Same procedure as d.

    h. V = Vo+g*t = 40 - 9.8*6 = -18.8 m/s.
    The negative sign means the stone is falling.

    i. h = ho - (V^2-Vo^2)/2g
    h = 161.6 - (50^2-0)/19.6 = 34 m.

    j. V=Vo + g*Tf = 0 + 9.8*5.74=56.3 m/s








    posted by Henry

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