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

Answer 1

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.

Answer 2

Answer: the equation you would use is

v = v0 +at

Explanation: