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justinbaker
Sep10-04, 02:32 PM
A football is kicked at ground level with a speed of 10.0 m/s at an angle of 39.0° to the horizontal. How much later does it hit the ground?

i found the Voy to be 10sin(39), then i used Y=Yo + Vot + .5at^2 and i solved for t. I found t to be 1.97s

but for some reason it is showing that i am wrong on my online hw, please help

Muzza
Sep10-04, 02:45 PM
Your calculator is working in radians, but you're working in degrees.

justinbaker
Sep10-04, 02:47 PM
thanks man, i feel like an idiot, your awsome

COCoNuT
Sep10-04, 11:42 PM
how did you set it up?

Y=Yo + Vot + .5at^2
0 = Yo + (10)t + .5(-9.8)t^2 <--a is -9.8 right? cause it's going down

what is Yo?


"i found the Voy to be 10sin(39)"

isnt that the velocity? so you cant plug that in for Yo.


how did you get your answer? i have the same type of problem

Pyrrhus
Sep11-04, 01:33 PM
A football is kicked at ground level with a speed of 10.0 m/s at an angle of 39.0° to the horizontal. How much later does it hit the ground?

Info:
V_{o} = 10 m/s
\theta_{o} = 39^o
Y_{o} = 0 The Football was at ground level
T_{f} = ? <- Flight time

Remember we can work this problem with V_{o} components
V_{yo} = V_{o}\sin(\theta_{o})
V_{xo} = V_{o}\cos(\theta_{o})

We can use the equation you used.
Y = 0 when it hits the ground.
0= Y_{o} + V_{yo}t_{f} - \frac{1}{2}gt^2_{f}
0= Y_{o} + V_{o}\sin(\theta_{o})t_{f} - \frac{1}{2}gt^2_{f}
0 = V_{o}\sin(\theta_{o})t_{f} - \frac{1}{2}gt^2_{f}

Simply solve for Flight Time.