A person throws a ball upward into the air with an initial velocity of 15.0 m/s.

Calculate the following:
a.) how high it goes.

b.) how long the ball is in the air before it comes back to his hand.

c.) at what time(s) will the ball pass a point 8.00 m.
If you got multiples times, are both solutions valid? Why or why not?

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To solve these questions, we can use the equations of motion for projectile motion. These equations can be derived using basic principles of physics, such as kinematics and the properties of gravity.

a.) To calculate how high the ball goes, we can use the equation for the maximum height reached by a projectile:

\[ h = \left(\frac{v^2}{2g}\right) \]

Where:
- \( h \) is the maximum height
- \( v \) is the initial velocity of the ball
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s²)

Substituting the given values into the equation:

\[ h = \left(\frac{15.0^2}{2 \times 9.8}\right) \]

Calculating this expression, we find:

\[ h \approx 11.47 \, \text{m} \]

Therefore, the ball reaches a maximum height of approximately 11.47 meters.

b.) The time the ball is in the air before it comes back to the hand can be determined using the equation for time of flight:

\[ T = \frac{2v}{g} \]

Where:
- \( T \) is the time of flight, the total time the ball is in the air

Substituting the given values:

\[ T = \frac{2 \times 15.0}{9.8} \]

Calculating this expression, we find:

\[ T \approx 3.06 \, \text{s} \]

Therefore, the ball is in the air for approximately 3.06 seconds.

c.) To find the time(s) when the ball passes a point 8.00 meters, we can use the equation of motion for displacement:

\[ h = vt - \frac{1}{2}gt^2 \]

Rearranging the equation, we get:

\[ t^2 - \frac{2v}{g}t + \frac{2h}{g} = 0 \]

Substituting the given values:

\[ t^2 - \frac{2 \times 15.0}{9.8}t + \frac{2 \times 8.00}{9.8} = 0 \]

Solving this quadratic equation, we find two solutions for \( t \):

\[ t_1 \approx 0.88 \, \text{s} \]
\[ t_2 \approx 5.54 \, \text{s} \]

Therefore, the ball passes the point at approximately 0.88 seconds and 5.54 seconds.

Both solutions are valid because the equation is quadratic, so it can have two real roots. The ball passes the specified point twice: once on the way up and once on the way back down. Therefore, both times represent valid instances when the ball passes the point.