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A stone is projected upward at an angle of 30 degree to the horizontal from the top of a tower 100m and it hits the ground at a point q. If the initial veloity of projection is 100ms cal the maximum height of the stone above the groud

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5 answers
  1. Vo = 100m/s[30o].
    Yo = 100*sin30 = 50 m/s.

    Y^2 = Yo^2 + 2g*h.
    h = (Y^2-Yo^2)/2g + ho.
    y = 0.
    g = -9.8 m/s^2.
    ho = 100 m.
    h = ?

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  2. 23

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  3. The vertical component of the velocity is needed. Therefore
    from v=u + gt
    0=usintita +gt
    100sin30 = 10t
    t=5s.
    so it takes 5 sec to get to the max height when thrown from the top of the building. To find the additional distance(height) travel we use this:
    S= ut + 1/2gt2.
    S= 0 + 1/2*10*25
    S= 125m.
    Therefore total height is 100+125=225m.

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  4. I hope that piece above was helpful

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  5. How do you solve it

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