The height $y$ of the ball after $t$ seconds is given by the equation:

$y = 44t - 16t^2$

This equation represents a quadratic function, where:

• $44t$ is the initial upward velocity component.
• $-16t^2$ accounts for the deceleration due to gravity.

Key Points:

1. Initial Height: When $t = 0$, the height $y$ is: $y(0) = 44(0) - 16(0)^2 = 0$ The ball starts from a height of 0 feet.

2. Maximum Height: To find the maximum height, we need to find the time when the velocity $v(t)$ becomes zero. The velocity is the derivative of the height function: $v(t) = \frac{dy}{dt} = 44 - 32t$ Setting $v(t) = 0$ to find the time at which the maximum height occurs: $44 - 32t = 0 \quad \Rightarrow \quad t = \frac{44}{32} = 1.375 \text{ seconds}$ Now, plug this value back into the height equation to find the maximum height: $y(1.375) = 44(1.375) - 16(1.375)^2 = 30.25 \text{ feet}$

3. Time to Hit the Ground: The ball will hit the ground when the height $y$ is zero: $44t - 16t^2 = 0 \quad \Rightarrow \quad t(44 - 16t) = 0$ This gives $t = 0$ (initial time) and $t = \frac{44}{16} = 2.75$ seconds when the ball hits the ground.

Summary:

• The ball reaches a maximum height of 30.25 feet at 1.375 seconds.
• It returns to the ground at 2.75 seconds.

Would you like more details on any of these calculations?

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Relative Questions:

1. How do you derive the maximum height formula for a quadratic function?
2. What is the significance of the derivative in finding the velocity?
3. Can you explain why the time to reach maximum height is exactly halfway to the time the ball hits the ground?
4. How would the equation change if the planet's gravity were different?
5. What happens to the trajectory if the initial velocity is increased?

Tip: When solving quadratic equations in physics, always consider the physical meaning of each root, as one may be non-physical (like a negative time).