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2 February, 01:57

Two small frogs simultaneously leap straight up from a lily pad. frog a leaps with an initial velocity of 0.551 m/s, while frog b leaps with an initial velocity of 1.75 m/s. when frog a lands back on the lily pad, what is the position and velocity of frog b? take upwards to be positive, and let the position of the lily pad be zero.

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  1. 2 February, 02:07
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    Define

    g = 9.8 m/s², acceleration due to gravity, positive downward.

    Assume that wind resistance may be neglected.

    Frog A:

    u = 0.551 m/s, launch velocity, upward.

    When the frog lands back on the pad, its vertical position is zero, and its vertical velocity will be 0.551 m/s downward.

    If the time of flight is t, then

    (0.551 m/s) * (t s) - 0.5 * (9.8 m/s²) * (t s) ² = 0

    0.551t - 4.9t² = 0

    t = 0, or t = 0.1124 s

    t = 0 corresponds to launch, and t = 0.1124 s corresponds to landing.

    Frog B:

    Launch velocity is 1.75 m/s

    When t = 0.1124 s, the position of the frog is

    s = (1.75 m/s) (0.1124 s) - 0.5 * (9.8 m/s²) * (0.1124 s) ²

    = 0.135 m

    The velocity of frog B is

    v = (1.75 m/s) - (9.8 m/s²) * (0.1124 s)

    = 0.6485 m/s

    Answer:

    When frog A lands on the ground,

    Frog B is 0.135 m above ground and its velocity is 0.649 m/s upward.
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