Finding the Height of a Muffin Throw

How can we determine the height at which the muffin was caught?

Given the initial and final velocities of the muffin, how can we calculate the height at which it was caught?

Calculating the Height of the Muffin

To find the height at which the muffin was caught, we can use the following kinematic equation for vertical motion:

[tex]v_f^2 = v_i^2 + 2as[/tex]

Where:

- [tex]v_f[/tex] is the final velocity (2.4 m/s upward)

- [tex]v_i[/tex] is the initial velocity (10.55 m/s upward)

- [tex]a[/tex] is the acceleration due to gravity (-9.81 m/s², negative because it acts downward)

- [tex]s[/tex] is the displacement (height above the initial point)

First, let's rearrange the equation to solve for [tex]s[/tex]:

[tex]s = \frac{v_f^2 - v_i^2}{2a}[/tex]

Now, we can plug in the values:

[tex]s = \frac{(2.4^2 - 10.55^2)}{2(-9.81)}[/tex]

Calculating this expression:

[tex]s \approx 6.23 \, \text{meters}[/tex]

So, the muffin was caught at a height of approximately 6.23 meters above the initial point, which was 1.50 meters above ground level. Therefore, the height above ground level at which the muffin was caught is 6.23 + 1.50 = 7.73 meters.