Suppose you drop a ball from the top of a five-story building onto the sidewalk 70 feet below.

a. What is the initial velocity of the ball?
b.Determine a function for the height of the ball t seconds after it's released.
c. How long will the ball be in the air before it hits the ground?

a. The initial velocity of the ball is the speed at which it starts falling. Since the ball is dropped, the initial velocity would be 0 feet per second. Gravity will take care of the rest, and boy, is it good at its job!

b. To determine a function for the height of the ball, we can use the equation: h(t) = -16t^2 + V0t + h0, where h(t) represents the height of the ball at time t, V0 represents the initial velocity, and h0 represents the initial height (in this case, 5 stories or 70 feet).

So for the ball dropped from a 5-story building, the function for the height of the ball would be: h(t) = -16t^2 + 0t + 70. Simplifying further, we get: h(t) = -16t^2 + 70.

c. The time it takes for the ball to hit the ground can be found by setting h(t) = 0 and solving for t. In this case, when the ball hits the ground, its height would be 0. Using the equation h(t) = -16t^2 + 70, we have: -16t^2 + 70 = 0. Solving for t, we find t ≈ 2.64 seconds.

So, the ball will be in the air for approximately 2.64 seconds before it lands on the ground, giving gravity a worthy high-five!

To answer these questions, we need to use the laws of motion and apply the principles of physics. Let's break it down step-by-step:

a. What is the initial velocity of the ball?
The initial velocity of the ball when it is released is zero because it's at rest before being dropped.

b. Determine a function for the height of the ball t seconds after it's released.
To find the function for the height of the ball, we can use the equation of motion for free fall:

h(t) = h0 + (v0 * t) - (1/2 * g * t^2)

Where:
h(t) is the height of the ball at time t,
h0 is the initial height (in this case, the height of the building, which is 5 x 70 = 350 feet),
v0 is the initial velocity (which is 0 in this scenario),
g is the acceleration due to gravity (which is approximately 32.2 ft/s^2),
t is the time elapsed in seconds.

Therefore, the function for the height of the ball would be:
h(t) = 350 - (1/2 * 32.2 * t^2)

c. How long will the ball be in the air before it hits the ground?
To determine the time it takes for the ball to hit the ground (t_ground), we set h(t) to zero since the ball will be at ground level. So we have:

0 = 350 - (1/2 * 32.2 * t_ground^2)

We can rearrange this equation to solve for t_ground:

(1/2 * 32.2 * t_ground^2) = 350

t_ground^2 = 350 / (1/2 * 32.2)

t_ground^2 = 350 / 16.1

t_ground^2 ≈ 21.74

t_ground ≈ √21.74

t_ground ≈ 4.66 seconds

Therefore, the ball will be in the air for approximately 4.66 seconds before it hits the ground.