You toss a rock up vertically at an initial speed of 34 feet per second and release it at an initial height of 6 feet. The rock will remain in the air for how many seconds?

Ah, the age-old question of how long a rock will stay airborne. Well, buckle up because we're about to go on a gravity-defying journey! Now, let's do some calculations while keeping things light and entertaining, shall we?

First, we need to find out when the rock hits the ground. Now, I'm no fortune-teller, but I can tell you a little secret: when a rock reaches its highest point, its velocity becomes zero. So, to find the time it takes for the rock to reach its highest point, we'll use a little physics trickery.

The initial velocity of the rock is 34 feet per second, and the acceleration due to gravity is about 32 feet per second squared. Using some math magic, we can find that it takes approximately 1 second for the rock to reach its highest point.

But wait, there's more! Now we need to determine the total time the rock spends in the air. Since we know it takes 1 second to reach the highest point, we can use symmetry to determine that it will also take 1 second to fall back down. This means the rock will spend a total of 2 seconds in the air.

So, my friend, if my calculations are as accurate as my rubber nose, the rock will remain in the air for a hilarious total of 2 seconds! Enjoy the show, and remember to always keep it rockin'!

To find out how long the rock will remain in the air, we can use the kinematic equation for vertical motion:

h = h0 + v0t - (1/2)gt^2

Where:
h = final height
h0 = initial height
v0 = initial velocity
g = acceleration due to gravity
t = time

In this case:
h0 = 6 feet (the initial height)
v0 = 34 feet per second (the initial velocity)
g = 32 feet per second squared (acceleration due to gravity)

We want to find the time (t) when the final height (h) is 0 (since the rock will touch the ground).

So, we substitute the values into the equation and solve for t:

0 = 6 + (34)t - (1/2)(32)(t^2)

Rearranging the equation gives:
0 = -16t^2 + 34t + 6

To solve this quadratic equation, we can either factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values for a, b, and c from the quadratic equation:

t = ( -34 ± √(34^2 - 4(-16)(6))) / (2(-16))
t = ( -34 ± √(1156 + 384)) / (-32)
t = ( -34 ± √(1540)) / (-32)

Calculating the square root gives:
t = ( -34 ± √(1540)) / (-32)
t ≈ -0.227s or t ≈ 2.227s

Since time cannot be negative, the rock will remain in the air for approximately 2.227 seconds.

To determine how long the rock remains in the air, we need to calculate the time it takes for the rock to reach its highest point and then fall back to its initial height.

First, let's find the time it takes for the rock to reach its highest point. The initial vertical velocity (upwards) is 34 feet per second, and the acceleration due to gravity is approximately -32.2 feet per second squared (assuming we are neglecting air resistance).

We can use the kinematic equation:

v = u + at

Where:
v is the final velocity (which is zero at the highest point),
u is the initial velocity,
a is the acceleration, and
t is the time taken.

Since we know the initial speed (u = 34 feet per second), the final velocity is zero, and the acceleration is -32.2 feet per second squared, we can rearrange the equation to solve for time:

t = (v - u) / a

t = (0 - 34) / -32.2
t ≈ 1.056 seconds (rounded to three decimal places)

Next, we need to calculate the time it takes for the rock to fall back to its initial height. Since the downward motion is analogous to a free-falling object under gravity, it will take the same amount of time to fall as it took to rise.

Therefore, the rock will remain in the air for approximately 2.112 seconds (twice the time it takes to reach the highest point) since it took about 1.056 seconds to rise and the same amount of time to fall.

h(t) = 6 + 34t - 16t^2

h=0 when t = 1/16 (17 + √385) = 2.29 seconds