Barbara serves the volleyball to Rhonda with an upward velocity of 22.5ft/s

22.5

ft
/
s
. The ball is 3.5
3.5
feet above the ground when she strikes it. How long does Rhonda have to react, before the volleyball hits the ground? Round your answer to two decimal places.

find t when h(t) = 0

since that is when it hits the ground.
h(t) = 3.5 + 22.5t - 16t^2

and geez, type fractions normally, wouldya?
as in 22.5 ft/s

Well, Rhonda certainly doesn't want to drop the ball on this one! Let's calculate how much time she has to react.

We can use the equation h = h0 + v0t - 16t^2, where h is the height, h0 is the initial height, v0 is the initial velocity, and t is the time.

Given that the initial height is 3.5 feet and the initial velocity is 22.5 ft/s, we can plug those values into the equation to get 3.5 = 22.5t - 16t^2.

Now, let's solve for t. Subtract 3.5 from both sides and rearrange the equation to get -16t^2 + 22.5t - 3.5 = 0.

Using the quadratic formula, t = (-22.5 ± sqrt(22.5^2 - 4(-16)(-3.5))) / (2(-16)).

After crunching the numbers, we get two possible solutions: t = 0.152 seconds and t = 1.039 seconds.

Now, since we're looking for the time it takes for the ball to hit the ground, we'll use the positive solution of t = 1.039s.

So, Rhonda has approximately 1.04 seconds to react before the volleyball lands. Better get those reflexes ready, Rhonda!

To find the time it takes for the volleyball to hit the ground, we can use the equation for vertical motion:

h = v₀t + 0.5gt²

where:
h = height above the ground (negative when below the ground)
v₀ = initial velocity (upward velocity in this case)
g = acceleration due to gravity (approximately -32.2 ft/s², negative because it acts downward)
t = time

In this case, the initial velocity (v₀) is 22.5 ft/s, the height (h) is -3.5 ft, and we want to find the time (t).

Plugging in the values:

-3.5 = (22.5)t + 0.5(-32.2)t²

Rearranging the equation:

0.5(-32.2)t² + (22.5)t - 3.5 = 0

This is a quadratic equation. We can solve it using the quadratic formula:

t = (-b ± √(b² - 4ac)) / (2a)

where a, b, and c are the coefficients of the equation.

In this case:
a = 0.5(-32.2) = -16.1
b = 22.5
c = -3.5

Plugging in the values:

t = (-(22.5) ± √((22.5)² - 4(-16.1)(-3.5))) / (2(-16.1))

Simplifying the equation:

t = (-22.5 ± √(506.25 - 225.4)) / (-32.2)

t = (-22.5 ± √(280.85)) / (-32.2)

Calculating the square root:

t = (-22.5 ± 16.77) / (-32.2)

Simplifying the division:

t = (-22.5 + 16.77) / (-32.2) or t = (-22.5 - 16.77) / (-32.2)

Calculating the values:

t = (-5.73) / (-32.2) or t = (-39.27) / (-32.2)

t ≈ 0.178 seconds or t ≈ 1.22 seconds

Rhonda has approximately 0.18 seconds or 1.22 seconds to react before the volleyball hits the ground.

To find the time it takes for the volleyball to hit the ground, we can use the kinematic equation:

𝑑 = 𝑣₀𝑡 + 1/2𝑎𝑡²

Where:
𝑑 = distance traveled (in this case, the initial height of the ball, 3.5 feet)
𝑣₀ = initial velocity (22.5 ft/s)
𝑎 = acceleration (from gravity, -32.2 ft/s², negative because it acts downward)
𝑡 = time

We can rearrange the equation to solve for 𝑡:

1/2𝑎𝑡² + 𝑣₀𝑡 - 𝑑 = 0

This is a quadratic equation with the form 𝑎𝑡² + 𝑏𝑡 + 𝑐 = 0, where 𝑎 = 1/2𝑎, 𝑏 = 𝑣₀, and 𝑐 = -𝑑. We can use the quadratic formula to find the solutions for 𝑡:

𝑡 = (-𝑏 ± √(𝑏² - 4𝑎𝑐)) / 2𝑎

Plugging in the values:

𝑡 = (-(22.5) ± √((22.5)² - 4(1/2𝑎)(-3.5))) / 2(1/2𝑎)

Simplifying:

𝑡 = (-22.5 ± √((22.5)² + 56)) / (-1)

Now we can calculate 𝑡.