Illustrate a detailed action scene where a bicyclist, with an athletic build and clad in green protective gear, is riding his black mountain bike at a fast speed off a flat roof. The building on which he does this stunt is painted blue and stands 3.6 meters high. Below, an expansive park waits for the biker’s daring landing. Capture the moment right as the biker takes off from the roof, so the distance between the edge of the building and the prospective landing spot remains a mystery.

a bicyclist rides off a flat roof at 13.4m/s, the roof is 3.6m above the ground, how far from the edge of the building does the bicyclist land? round the answer to the nearest meter.

Vo=13.4m/s

y=-3.6m (i really don't now if this should be negative, negative beacuse it says above the ground😅)
g=9.8m/s^2
0=means degree(if the problem didn't give a degree given it automatic zero)
Vertical and Horizontal Displacement Relationship Formula

y=xtan0 - gx^2/
2Vo^2cos^20

Formula
Vo=√gx^2 /
(xtan0-y)(2cos^20)

Substitute/Place the given
13.4m/s=√(9.8m/s^2) x^2 /
(xtan0- (-3.6m))(2cos^20)

Multiply
13.4m/s=√9.8m/s^2 x^2 /
(3.6m)(2)

Divide
13.4m/s=√9.8m/s^2 x^2 /
7.2m

Square root and squared (so square root and squared of x gone)
13.4m/s=√1.381111111 s^2 x^2

Divide both sides to left the x
13.4m/s / 1.166666667 s=1.166666667 s^2 x / 1.166666667 s

11.48571428 m = x
11 m or 11.49m = x

I hope you understand hehehe

h = Vo*t + 0.5g*t^2. = 3.6

0 * 4.9t^2 = 3.6
t^2 = 0.735 .
Tf = 0.857 s. = Fall time.

d = Xo * Tf = 13.4m/s * 0.857s. = 11 m.

Well, let's calculate it "bike-nically" for you! When our brave cyclist takes a leap of faith, the horizontal velocity remains constant. Therefore, we can ignore that component for now. We are only concerned with the vertical motion.

Now, using the kinematic equation for vertical motion, we have:

vf^2 = vi^2 + 2ad

Where:
vf = final velocity (0 m/s at the top)
vi = initial velocity (13.4 m/s downward)
a = acceleration due to gravity (-9.8 m/s^2, because gravity is always a downer)
d = distance traveled vertically (3.6 m downward)

Rearranging the equation:

0 = (13.4 m/s)^2 + 2(-9.8 m/s^2)d

Simplifying further:

0 = 179.56 m^2/s^2 - 19.6 m/s^2 * d

19.6 m/s^2 * d = 179.56 m^2/s^2
d = 179.56 m^2/s^2 / 19.6 m/s^2
d ≈ 9.16 m

So, the intrepid rider lands approximately 9 meters from the edge of the building. I hope they remembered to bring their parachute! 🚲🪂

To find the distance from the edge of the building where the bicyclist lands, we need to use the equations of motion. The equation we will use is:

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

where:
S = distance traveled (we need to find this)
u = initial velocity (13.4 m/s)
t = time taken to reach the ground (we need to find this)
a = acceleration due to gravity (-9.8 m/s^2)

First, let's find the time it takes for the bicyclist to reach the ground. We will use the equation:

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

Rearranging the equation to solve for t:

t^2 + 2u/a * t - 2S/a = 0

Substituting the given values:
u = 13.4 m/s
a = -9.8 m/s^2 (negative because it acts in the opposite direction of motion)
S = 3.6 m

t^2 + 2 * 13.4 / (-9.8) * t - 2 * 3.6 / (-9.8) = 0

Let's solve this quadratic equation to find t.

Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a

Substituting the values:
a = 1
b = 2 * 13.4 / (-9.8)
c = -2 * 3.6 / (-9.8)

t = (-(2 * 13.4 / (-9.8)) ± √((2 * 13.4 / (-9.8))^2 - 4 * 1 * (-2 * 3.6 / (-9.8)))) / 2 * 1

Calculate this to find the value of t.

To find the distance from the edge of the building where the bicyclist lands, we can use the kinematic equation for horizontal motion:

\[ \text{Distance} = \text{Horizontal Velocity} \times \text{Time} \]

First, we need to calculate the time it takes for the person to fall from the roof to the ground. To do this, we can use the kinematic equation for vertical motion:

\[ \text{Vertical Displacement} = \text{Initial Vertical Velocity} \times \text{Time} + \frac{1}{2} \times \text{Vertical Acceleration} \times \text{Time}^2 \]

The vertical displacement is the difference in height between the roof and the ground, which is 3.6 meters. The initial vertical velocity is 0 m/s (since the person starts from rest vertically), and the vertical acceleration is due to gravity and is approximately 9.8 m/s². Solving for time:

\[ 3.6 = 0 \times \text{Time} + \frac{1}{2} \times 9.8 \times \text{Time}^2 \]

Rearranging the equation, we get:

\[ 4.9 \times \text{Time}^2 = 3.6 \]

Dividing both sides of the equation by 4.9:

\[ \text{Time}^2 = \frac{3.6}{4.9} \]

Taking the square root of both sides:

\[ \text{Time} = \sqrt{\frac{3.6}{4.9}} \]

Now, we can calculate the horizontal distance traveled using the formula we mentioned earlier:

\[ \text{Distance} = \text{Horizontal Velocity} \times \text{Time} \]

Substituting the given values, the horizontal velocity is 13.4 m/s, and the time is the value we just calculated. Evaluating the expression:

\[ \text{Distance} = 13.4 \times \sqrt{\frac{3.6}{4.9}} \]

Calculating this, we get:

\[ \text{Distance} \approx 11 \, \text{meters} \]

Therefore, the bicyclist lands approximately 11 meters away from the edge of the building.