A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the vertical velocity of the frog at t

Question

A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the vertical velocity of the frog at the moment that it left the ground. Ignore air resistance. Which kinematic formula would be used

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Bình An 4 years 2021-08-15T11:44:26+00:00 2 Answers 111 views 0

Answers ( )

    0
    2021-08-15T11:45:37+00:00

    Answer:

    You would use the kinematic formula:

        \Delta y=V_{0y}\times t-g\times t^2/2

    Explanation:

    The upwards vertical motion is ruled by the equation:

            y=y_0+V_{0y}\times t-g\times t^2/2

    Where:

           y \text{ is the position at the time }t:y=0.1m

           y_0\text{ is the initial position: }y_0=0

           t=2s

           g\text{ is the gravitational acceleration: }\approx 9.8m/s^2

           V_{0y}\text{ is the initial vertical velocity}

    Naming Δy = y – y₀, the equation becomes:

          \Delta y=V_{0y}\times t-g\times t^2/2

    Then, you just need to substitute with Δy = 0.1m, t = 2s, and g = 9.8m/s², ans solve for the intital vertical velocity.

    0
    2021-08-15T11:46:08+00:00

    Answer: the equation you would use is

    v = v0 +at

    Explanation:

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