A track star in the long jump goes into the jump at 12 m/s and launches herself at 20 degrees above the horizontal. How long is she in the air before returning to Earth? (g= 9.81 m/s squared)
I tried using some formulas but I'm using all of the wrong formulas for this question. I can work this out if someone can give me the correct formula to use. :)
Thanks!
You know the intial height, and the final height. You know the initial velocity.
hfinal=hinitial + v*sin20*t -1/2 g t^2
solve for t.
Thanks! ^^
200
0.83 seconds
To determine how long the track star is in the air before she returns to Earth, we need to use the kinematic equations. Specifically, we will use the equation for vertical motion, as the track star launches herself at an angle above the horizontal.
The equation we will use is:
y = y0 + v0y * t - (1/2) * g * t^2,
where:
- y is the vertical displacement from the starting point (y = 0 when the track star takes off and y < 0 when she returns to Earth),
- y0 is the initial vertical position (y0 = 0 in this case),
- v0y is the initial vertical component of velocity (v0y = v0 * sin(theta)),
- t is the time,
- g is the acceleration due to gravity (g = 9.81 m/s^2).
Since she launches herself at 20 degrees above the horizontal, we can convert this angle to radians and calculate the value of v0y:
theta = 20 degrees = 20 * pi / 180 radians ≈ 0.3491 radians,
v0y = v0 * sin(theta) = 12 m/s * sin(0.3491) ≈ 6.8956 m/s.
Now, let's plug in the known values and solve for the time:
y = 0 + 6.8956 * t - (1/2) * 9.81 * t^2.
In this equation, y = 0 when the track star lands back on the ground. To find the time she spent in the air, we need to solve for t when y = 0. Rearrange the equation to:
(1/2) * 9.81 * t^2 - 6.8956 * t = 0.
This is a quadratic equation, and we can solve it using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a).
In our case, a = (1/2) * 9.81, b = -6.8956, and c = 0. Plugging in these values, we get:
t = (-(-6.8956) ± sqrt((-6.8956)^2 - 4 * (1/2) * 9.81 * 0)) / (2 * (1/2) * 9.81).
Simplifying further:
t = (6.8956 ± sqrt(47.54132)) / 4.905.
Calculating the square root:
t ≈ (6.8956 ± 6.899) / 4.905.
Now, we have two possible solutions:
t1 ≈ (6.8956 + 6.899) / 4.905 ≈ 2.813 seconds,
t2 ≈ (6.8956 - 6.899) / 4.905 ≈ -0.409 seconds.
Since time cannot be negative in this context, we discard the negative solution. Therefore, the track star is in the air for approximately 2.813 seconds before returning to Earth.