5) After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 22.0 m/s when it reaches a maximum height of 9.0 m above the ground.

What is the speed of the ball when it leaves Sarah's hand?

Where does the t=v/9.8 come form

While Steve is mostly right there is an error in his algebra. The third line should read:

9.0=-4.9(v^2/9.8^2)+v(v/9.8)+1.5.
The rest can be correctly solved using this equation.

h(t) = -4.9t^2 + vt + 1.5

max height reached when t = v/9.8
9.0 = -4.9(v^2/4.9^2) + v(v/4.9) + 1.5
v = 7√3 = 12.1244 m/s

That is the vertical component. So, the speed is

√(7√3)^2 + 22^2 = 25.12 m/s

2*a(x-x0)=(v^2-v0^2)

2*-9.8(9-1.5)=(v^2-(22*22))
v=18.35

To find the speed of the ball when it leaves Sarah's hand, we can use the principle of conservation of mechanical energy.

The principle of conservation of mechanical energy states that the total mechanical energy of a system, which includes both kinetic energy and potential energy, remains constant as long as no external forces are acting on the system.

In this case, when the ball reaches its maximum height, it has only potential energy and no kinetic energy. Therefore, the mechanical energy of the ball when it leaves Sarah's hand is equal to the potential energy at its maximum height.

The potential energy is given by the formula:

Potential energy = mass * gravity * height

Here, the height is 9.0 m above the ground. We can assume that the mass of the ball is not given, but since mass cancels out when calculating speed, we don't need it.

The potential energy at the maximum height is:

Potential energy = mass * gravity * height
= 9.8 m/s^2 * 9.0 m

Now, to find the speed when the ball leaves Sarah's hand, we need to find the total mechanical energy of the ball at that point. This is the sum of the potential energy and the kinetic energy.

The kinetic energy of an object is given by the formula:

Kinetic energy = 0.5 * mass * velocity^2

Since we don't know the mass of the ball, we'll use a variable, "m," to represent it. We can see that the mass will cancel out in our calculations, so it doesn't affect the final answer.

At the top of the ball's trajectory (when it reaches its maximum height), the kinetic energy is zero. Therefore, the total mechanical energy is equal to the potential energy at that point.

Therefore, we can set up the equation:

Potential energy = Total mechanical energy
= 0.5 * mass * velocity^2 + mass * gravity * height

Now, we can plug in the values given in the question:
Potential energy = 9.8 m/s^2 * 9.0 m
= 88.2 J

Therefore, we have:
88.2 J = 0.5 * mass * velocity^2 + mass * 9.8 m/s^2 * 1.5 m

Simplifying further:

88.2 J = 0.5 * mass * velocity^2 + 14.7 * mass

Since mass is a common factor, we can factor it out:

88.2 J = mass * (0.5 * velocity^2 + 14.7)

Now, we can isolate the term for velocity:

0.5 * velocity^2 = (88.2 J - 14.7 * mass) / mass

Dividing both sides by 0.5:

velocity^2 = [(88.2 J - 14.7 * mass) / mass] * 2

Finally, taking the square root of both sides to solve for velocity:

velocity = sqrt([(88.2 J - 14.7 * mass) / mass] * 2)

Since the mass of the ball is not provided, we cannot calculate the exact speed of the ball when it leaves Sarah's hand. However, this equation shows the relationship between mass and velocity.