A track star in the long jump goes into the jump at 12 m/s and launches herself at 20.0° above the horizontal. How long is she in the air before returning to Earth? (g = 9.81 m/s2)

Vo = 12m/s[20o]

Xo = 12*cos20 = 11.28 m/s.
Yo = 12*sin20 = 4.10 m/s.

Y = Yo + g*Tr = 0 @ max. ht.
4.10 - 9.8Tr = 0
-9.8Tr = -4.10
Tr = 0.419 s. = Rise time.

Tf = Tr = 0.419 s.

T = Tr + Tf = 0.419 + 0.419 = 0.838 s.=
Time in air.

13m

superb

Displacement is 9.24m

Well, let's do some calculations and see how long our track star friend spends in the air. If she goes into the jump at 12 m/s, we can break that down into its vertical and horizontal components. The horizontal component will remain constant throughout the jump because there's no force acting on it.

Now, let's focus on the vertical component. The initial velocity in the vertical direction can be found by multiplying the initial velocity (12 m/s) by the sine of the launch angle (20.0°). So, V_y = 12 m/s * sin(20.0°) = 4.09 m/s.

The time it takes for an object to reach its peak height can be found using the following equation: V_f = V_i + gt, where V_f is the final velocity (which is zero when the object reaches its peak), V_i is the initial velocity, g is the acceleration due to gravity (-9.81 m/s²), and t is the time.

Since the final velocity is zero at the peak of the jump, we can plug in the values and solve for t: 0 = 4.09 m/s - 9.81 m/s² * t. Solving for t yields t = 0.418 seconds.

Now, remember that the total time in the air is twice the time it takes to reach the peak height, so the track star will spend 2 * 0.418 s = 0.836 seconds in the air before returning to Earth.

So, our track star will have approximately 0.836 seconds to contemplate her life choices before landing back on solid ground.

To calculate the time the track star is in the air before returning to Earth, we can use the equations of motion and consider the vertical motion component separately.

Step 1: Calculate the initial vertical velocity (viy):
Given the launch angle, we can calculate the initial vertical velocity using the formula:
viy = vi * sin(theta)
where
vi = initial velocity (12 m/s)
theta = launch angle (20.0°)

Substituting the values:
viy = 12 m/s * sin(20.0°)
viy ≈ 4.09 m/s

Step 2: Calculate the time taken to reach the maximum height (t1):
To calculate the time taken to reach the maximum height, we need to consider the vertical motion component. The time taken to reach maximum height is equal to the time it takes for the vertical velocity to become zero.
Using the equation of motion:
0 = viy - g * t1
where
viy = initial vertical velocity (4.09 m/s)
g = acceleration due to gravity (9.81 m/s^2)
t1 = time taken to reach maximum height

Solving for t1:
t1 = viy / g
t1 = 4.09 m/s / 9.81 m/s^2
t1 ≈ 0.42 seconds

Step 3: Calculate the total time in the air (t):
To calculate the total time in the air, we need to consider the time taken to reach maximum height (t1). Since the motion has symmetry, the total time in the air is twice the time taken to reach maximum height.

t = 2 * t1
t ≈ 2 * 0.42 seconds
t ≈ 0.84 seconds

Therefore, the track star is in the air for approximately 0.84 seconds before returning to Earth.

To find out how long the track star is in the air before returning to Earth, we can use the kinematic equations of motion. The key equation we need to use is the equation for vertical motion.

The equation for vertical motion is given by:

s = ut + (1/2)gt²

Where:
s = vertical displacement (in this case, the height above the ground)
u = initial vertical velocity (in this case, the vertical component of the launch velocity)
t = time
g = acceleration due to gravity

First, we need to find the initial vertical velocity (u). We can calculate it using trigonometry. The initial vertical velocity can be found by multiplying the launch velocity (12 m/s) by the sine of the launch angle (20.0°).

u = 12 m/s x sin(20.0°)

Now we can substitute the known values into the equation for vertical motion to find the time (t) when the track star reaches the ground (s = 0):

0 = (12 m/s x sin(20.0°))t + (1/2)(9.81 m/s²)t²

Simplifying this equation, we get a quadratic equation in terms of t:

(1/2)(9.81 m/s²)t² + (12 m/s x sin(20.0°))t = 0

We can solve this equation to find the two possible values of t. One value will be the time when the track star is in the air, and the other value will be the time when she hits the ground.

Using a mathematical equation solver or quadratic formula, we can obtain the time when the track star hits the ground. The positive value of t would represent the time when she returns to Earth.

After calculating the time, you would have the answer to the question, "How long is she in the air before returning to Earth?"