# You throw a ball into the air with an initial vertical velocity of 31 feet per second. The ball leaves your hand when it is 6 feet above the ground. You catch the ball when it reaches a height of 4 feet. After how many seconds do you catch the ball?

## solve for t in

6 + 31t - 16t^2 = 4

t = 2

## Well, catching a ball can be quite the juggling act, especially with all these numbers involved! Let's see if we can clown around with some physics.

We know that the initial velocity of the ball is 31 feet per second, and it starts 6 feet above the ground. So, to find the time it takes to catch the ball, we'll need to know when it reaches a height of 4 feet.

Now, let's do some calculations (or circus tricks, if you will). The downward acceleration due to gravity is approximately 32 feet per second squared. So, we can use the equation of motion h = h₀ + v₀t - 16t² to solve for t.

Substituting the given values, we have:

4 = 6 + 31t - 16t²

Rearranging things a bit, we get:

16t² - 31t + 2 = 0

Now, let's apply some mathematical circus magic and use the quadratic formula:

t = (-b ± √(b² - 4ac)) / (2a)

Plugging the values from our equation, we have:

t = (-(-31) ± √((-31)² - 4 * 16 * 2)) / (2 * 16)

After some ridiculous calculations, we find that the positive solution is approximately 2.08 seconds.

So, after approximately 2.08 seconds, you'll catch that elusive ball! Just make sure to stay focused and not get distracted by all the clowning around.

## To calculate the time it takes for you to catch the ball, we can use the equation of motion for vertical displacement:

s = ut + (1/2)at^2

Where:

s = displacement (vertical position)

u = initial velocity

t = time

a = acceleration due to gravity (approximately -32.2 ft/s^2, since the ball is moving upward in this case)

We can use this equation to find the time it takes for the ball to reach 4 feet above the ground.

Let's solve it step-by-step:

Step 1: Convert the initial velocity to negative, since the ball is moving upward.

Initial velocity (u) = -31 ft/s

Step 2: Set the displacement (s) to 4 feet and solve for time (t).

s = ut + (1/2)at^2

4 = -31t + (1/2)(-32.2)t^2

Step 3: Rearrange the equation in standard quadratic form.

16.1t^2 - 31t + 4 = 0

Step 4: Solve the quadratic equation using the quadratic formula:

For a quadratic equation in the form ax^2 + bx + c = 0, the quadratic formula is given by:

t = (-b ± √(b^2 - 4ac)) / 2a

Applying the formula, we have:

t = (31 ± √((-31)^2 - 4(16.1)(4)) / 2(16.1)

Step 5: Calculate the discriminant (b^2 - 4ac) to determine the nature of the roots.

Discriminant (D) = (-31)^2 - 4(16.1)(4)

Step 6: Determine the nature of the roots based on the discriminant.

If D > 0, there are two real and distinct roots.

If D = 0, there are two real and equal roots.

If D < 0, there are two complex roots.

Step 7: Calculate the values of t using the quadratic formula.

t = (31 + √((-31)^2 - 4(16.1)(4)) / 2(16.1)

t = (31 - √((-31)^2 - 4(16.1)(4)) / 2(16.1)

Step 8: Simplify the solutions.

t = (31 + √(961 - 257.6)) / 32.2

t = (31 + √703.4) / 32.2

t = (31 - √(961 - 257.6)) / 32.2

t = (31 - √703.4) / 32.2

Step 9: Calculate the final values of t.

t ≈ 0.746 seconds

t ≈ 1.164 seconds

Therefore, you catch the ball at approximately 0.746 seconds and 1.164 seconds after throwing it into the air.

## To determine the time it takes for you to catch the ball, we can use the kinematic equation for vertical motion:

h = h₀ + v₀t - 0.5gt²

where:

h = height at a given time (in this case, 4 feet)

h₀ = initial height (6 feet above the ground)

v₀ = initial vertical velocity (31 feet per second)

t = time

g = acceleration due to gravity (approximately 32.2 feet per second squared)

We need to rearrange the equation to solve for t. Let's substitute the known values first:

4 = 6 + (31)t - 0.5(32.2)t²

Now, let's simplify the equation:

0 = -0.5(32.2)t² + 31t - 2

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b² - 4ac)) / 2a

In this case, a = -0.5(32.2), b = 31, and c = -2. Let's substitute these values into the formula:

t = (-(31) ± √((31)² - 4(-0.5(32.2))(-2))) / 2(-0.5(32.2))

Simplifying further:

t = (-31 ± √(961 + 128.8))/(32.2)

t = (-31 ± √1089.8)/32.2

Now, let's calculate the values within the square root:

t = (-31 ± √1089.8)/32.2

t ≈ (-31 ± 32.99)/32.2

This gives us two possible solutions:

t ≈ (-31 + 32.99)/32.2 ≈ 1.99/32.2 ≈ 0.06 seconds

t ≈ (-31 - 32.99)/32.2 ≈ -63.99/32.2 ≈ -1.99 seconds

Since time cannot be negative in this scenario, we can discard the negative solution. Therefore, you catch the ball approximately 0.06 seconds after it leaves your hand.