Let's solve this problem step by step.

We are given the position function for a ball thrown into the air:

$x(t) = -5t^2 + 3t + 2$

1. Velocity Function:

The velocity function $v(t)$ is the derivative of the position function $x(t)$. So, let's differentiate:

$v(t) = \frac{d}{dt} [-5t^2 + 3t + 2] = -10t + 3$

2. Acceleration Function:

The acceleration function $a(t)$ is the derivative of the velocity function:

$a(t) = \frac{d}{dt} [-10t + 3] = -10$

So, the acceleration is constant and equals $-10 \, \text{m/s}^2$, which is expected as it corresponds to the acceleration due to gravity.

3. Time when the ball reaches its highest point:

At the highest point, the velocity of the ball is 0, i.e., $v(t) = 0$. Set the velocity function to 0 and solve for $t$:

$-10t + 3 = 0$ $10t = 3$ $t = \frac{3}{10} = 0.3 \, \text{seconds}$

So, the ball reaches its highest point at $t = 0.3 \, \text{seconds}$.

4. When does the ball hit the ground?

The ball hits the ground when its position $x(t)$ is 0. Set $x(t) = 0$ and solve for $t$:

$-5t^2 + 3t + 2 = 0$

This is a quadratic equation, and we can solve it using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For $ax^2 + bx + c = 0$, we have:

• $a = -5$,
• $b = 3$,
• $c = 2$.

Now, plug in the values:

$t = \frac{-3 \pm \sqrt{3^2 - 4(-5)(2)}}{2(-5)} = \frac{-3 \pm \sqrt{9 + 40}}{-10} = \frac{-3 \pm \sqrt{49}}{-10}$

$t = \frac{-3 \pm 7}{-10}$

So, the two possible solutions for $t$ are:

$t = \frac{-3 + 7}{-10} = \frac{4}{-10} = -0.4 \quad (\text{not valid since time cannot be negative}),$

$t = \frac{-3 - 7}{-10} = \frac{-10}{-10} = 1 \, \text{second}.$

Thus, the ball hits the ground at $t = 1 \, \text{second}$.

5. How fast is the ball going when it hits the ground?

To find the speed when the ball hits the ground, we evaluate the velocity function at $t = 1$:

$v(1) = -10(1) + 3 = -10 + 3 = -7 \, \text{m/s}.$

The speed is the magnitude of the velocity, so the ball is moving at $7 \, \text{m/s}$ when it hits the ground.

Summary:

• The velocity function is $v(t) = -10t + 3$.
• The acceleration function is $a(t) = -10 \, \text{m/s}^2$.
• The ball reaches its highest point at $t = 0.3 \, \text{seconds}$.
• The ball hits the ground at $t = 1 \, \text{second}$.
• The ball is moving at $7 \, \text{m/s}$ when it hits the ground.

Do you need more details on any of these steps?

Here are 5 related questions you might explore:

1. What would happen if the initial velocity were doubled? How would that affect the maximum height and time to hit the ground?
2. How does the acceleration due to gravity affect objects of different masses in free fall?
3. Can you derive the position function for a ball thrown with an initial velocity of 5 m/s upward?
4. How would air resistance affect the motion of the ball in this problem?
5. How can we use energy conservation to verify the speed of the ball when it hits the ground?

Tip: Remember that velocity can be positive or negative, depending on the direction of motion, but speed is always a positive quantity!