To determine how long it will take for the ball to reach the ground, we can use the given formula:

$s(t) = -16t^2 + v_0t + s_0$

where:

• $v_0$ is the initial velocity (2020 feet per second),
• $s_0$ is the initial altitude (77 feet),
• $s(t)$ is the altitude at time $t$ (which will be 0 when the ball hits the ground).

We need to solve for $t$ when $s(t) = 0$.

The equation becomes:

$0 = -16t^2 + 2020t + 77$

This is a quadratic equation in the form $at^2 + bt + c = 0$, where $a = -16$, $b = 2020$, and $c = 77$. We can solve this quadratic equation using the quadratic formula:

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

Plugging in the values:

$t = \frac{-2020 \pm \sqrt{2020^2 - 4(-16)(77)}}{2(-16)}$

Let's calculate this step by step.

1. Calculate the discriminant: $\text{Discriminant} = 2020^2 - 4(-16)(77)$

2. Take the square root of the discriminant.

3. Use the quadratic formula to solve for $t$.

Let me compute this.The solutions for the time $t$ are approximately:

$t_1 \approx -0.038 \text{ seconds}$ $t_2 \approx 126.288 \text{ seconds}$

Since time cannot be negative, we discard $t_1$. Therefore, the ball will take approximately 126.29 seconds to reach the ground.

Would you like more details or have any questions?

Here are 5 related questions:

1. What would be the maximum height reached by the ball?
2. How would the time change if the initial velocity were doubled?
3. What if the ball were thrown from a higher altitude?
4. How would air resistance affect the time to reach the ground?
5. Can you determine the velocity of the ball just before it hits the ground?

Tip: When solving quadratic equations, always consider the physical context to determine which solution is meaningful.