Calculating Jump Time of a Star in the Long Jump

[Makaroon]

Homework Statement

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

Homework Equations

first I look for the v0(cos) and then I look for vfy= v0(sin)
I used vfy/g and I don't get the answer. I have the answer which is .83 s
but I can't get it on my own...

The Attempt at a Solution

I used vfy/g to get the time, but I don't get the correct answer I get .4

 

[ShawnD]

Homework Statement

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

You probably just forgot the downfall time. Here's what the setup is like for finding the half-time in the air.
0 = (m)(v) - (F)(t)
Add a sine function to represent the vertical velocity
0 = (m)(v)sin(theta) - (m)(g)(t)
factor out the mass and start filling in values
0 = (12)sin(20) - (9.8)(t)
t = 0.4188 going up
Add another 0.4188 for the going down to get a total of 0.8376 seconds.

 

[Makaroon]

thanks! yes I get it now, I was forgetting to multiply the time with the gravity.
thanks again. :)