A referee tosses a coin straight up with a velocity of 5.25 m/s, how high does it go above its point of release? (Hint: How fast is it moving at the maximum height?)

The vertical velocity is 0 at the top

v = Vi - 9.81 t

0 = 5.25 - 9.81 t

t = .535 second to the top

h = 0 + Vi t - 4.9 t^2

h = 5.25 (.535) - 4.9 (.535^2)

h = 1.40 meter

alternate and faster way:

(1/2) m v^2 = m g h
[ conservation of energy ]
h = v^2/2g = 5.25^2/19.6 = 1.40 meter

To determine how high the coin goes above its point of release, we need to use some basic physics principles.

First, let's address the hint provided. At the maximum height, the velocity of the coin will be zero momentarily before it starts falling back down. This occurs because the coin's motion undergoes a change from upward to downward. So, to find the maximum height, we need to calculate the time it takes for the coin to reach that point.

To begin, we need to understand the relationship between velocity, time, and acceleration. In this case, the acceleration acting on the coin is due to gravity, which is approximately 9.8 m/s² downward (assuming no air resistance).

Using the equation of motion:
vf = vi + at

Where:
vf = final velocity (0 m/s at the maximum height)
vi = initial velocity (5.25 m/s)
a = acceleration (-9.8 m/s²)
t = time

We can rearrange the equation to solve for time:
t = (vf - vi) / a

Substituting in the values we know:
t = (0 - 5.25) / -9.8

Calculating:
t ≈ 0.54 seconds

Now that we have the time taken to reach the maximum height, we can calculate the distance using the equation of motion:
h = vi*t + (1/2)*a*t^2

Where:
h = height
vi = initial velocity (5.25 m/s)
t = time (0.54 seconds)
a = acceleration (-9.8 m/s²)

Substituting the values:
h = 5.25 * 0.54 + (1/2) * (-9.8) * (0.54^2)

Calculating:
h ≈ 1.43 meters

Therefore, the coin goes approximately 1.43 meters above its point of release.