Anonymous
Anonymous asked in Science & MathematicsPhysics · 7 months ago

Velocity and displacement question?

An object is thrown straight up with a velocity, in ft/s, given by v(t)=−32t+65, where t is in seconds, from a height of 38 feet.

1) What is the object's maxiumum velocity?

2) What is the object's maximum displacement?

3) When is the object's displacement 0?

4) What is the object's maximum height?

2 Answers

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  • NCS
    Lv 7
    7 months ago
    Favorite Answer

    1) max velocity at launch -- 65 ft/s

    (agree with other post)

    2) In order for the answers to parts 2 and 4 to be different, the launch point must be considered the location of zero displacement.

    max s = v²/2g = 65²/64 = 66

    3) s(t) = ∫ v(t) dt = -16t² + 65t + C

    we're told s(0) = 38, so C = 38 and

    s(t) = -16t² + 65t + 38

    for displacement zero,

    s(0) = 0 = -16t² + 65t + 38

    which has a negative root we can ignore

    and a positive root at

    t ≈ 4.6

    4) 38 + 66 = 104

    Hope this helps!

  • Dylan
    Lv 6
    7 months ago

    Strictly decreasing so v(0) is max

    65 ft/s

    v(t) = 0, t = 65/32 s

    v'(t) = s(t) = -16t^2 + 65t

    Max at 65/32 s

    s(0) = 38

    Ans 2 + Ans 3

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