The problem you provided involves finding key points about the motion of a golf ball, modeled by the equation:

$h(t) = -16t^2 + 45t$

Where $h(t)$ represents the height (in feet) of the ball at time $t$ (in seconds).

Let’s go through each part of the question one by one.

a) How long does it take for the golf ball to reach its maximum height?

To find the time when the ball reaches its maximum height, we need to find the vertex of the parabola described by the equation. Since the equation is in the form $h(t) = at^2 + bt + c$, the time at which the maximum height occurs is given by:

$t = \frac{-b}{2a}$

Here, $a = -16$ and $b = 45$. Plugging these values into the formula:

$t = \frac{-45}{2(-16)} = \frac{-45}{-32} = 1.40625 \text{ seconds}$

b) What is its maximum height?

Now that we know the time at which the ball reaches its maximum height, we can substitute $t = 1.40625$ into the original height equation to find the maximum height:

$h(1.40625) = -16(1.40625)^2 + 45(1.40625)$

Let me calculate this value for you. Calculate the maximum height at t = 1.40625 t_max = 1.40625 h_max = -16 * t_max**2 + 45 * t_max h_maxThe maximum height of the golf ball is approximately 31.64 feet.

c) How long will it take for it to hit the ground?

The golf ball will hit the ground when the height $h(t) = 0$. To find this time, we solve the equation:

$-16t^2 + 45t = 0$

Factoring out $t$:

$t(-16t + 45) = 0$

Setting each factor equal to zero:

$t = 0 \quad \text{or} \quad -16t + 45 = 0$

Solving the second equation:

$t = \frac{45}{16} = 2.8125 \text{ seconds}$

So, the ball will hit the ground at 2.8125 seconds.

d) When will the golf ball be 10 feet above the ground?

To find when the ball is 10 feet above the ground, we set $h(t) = 10$ and solve:

$-16t^2 + 45t = 10$

Rearrange the equation:

$-16t^2 + 45t - 10 = 0$

Let me solve this quadratic equation for $t$. sympy as sp

Define the variable and the quadratic equation

t = sp.symbols('t') quadratic_eq = -16 * t**2 + 45 * t - 10

Solve the quadratic equation

solutions = sp.solve(quadratic_eq, t) solutionsThe solutions to the equation $-16t^2 + 45t - 10 = 0$ are:

$t = \frac{45}{32} - \frac{\sqrt{1385}}{32} \quad \text{and} \quad t = \frac{45}{32} + \frac{\sqrt{1385}}{32}$

Approximating these values:

$t_1 \approx 0.296 \quad \text{seconds} \quad \text{and} \quad t_2 \approx 3.544 \quad \text{seconds}$

So, the golf ball will be 10 feet above the ground at approximately 0.296 seconds and 3.544 seconds.

Would you like any additional details or clarifications? Here are some related questions for further exploration:

1. How do you interpret the meaning of the two times when the ball is 10 feet above the ground?
2. How can we determine the initial velocity and its impact on the trajectory?
3. What would happen if the coefficient of $t^2$ changed in the equation?
4. How does the maximum height correlate with the initial velocity in a projectile motion equation?
5. Can we use this same equation to model other objects with different initial velocities?

Tip: When dealing with projectile motion equations, the vertex of the parabola gives you the maximum or minimum values, which represent crucial points like maximum height or the point where the object hits the ground.