Posted by someone on Saturday, August 1, 2009 at 10:32am.
What is the vertical velocity component?
how long will it stay in the air?
hfinal=hinitial+Vv*timeinair-1/2 g timinair^2
solve for time in air.
Then horizontal distance=Vh*timeinair
I didn't get your answers.
What is the vertical velocity component?
that's 9 m/s right?
the time i got is 4.4/5.34s
which i think are wrong..
correct. I didn't check the quadreatic solutions, but it is slighly over 1.8 seconds
so which should i use 4.4 or 5.34?
which is slighty over 1.8seconds anyway? and shud i just subtract it?
Neither of your answers is correct. The shot returns to 2.2 m height after travelling
X = (V^2/g)sin(2*40) = 19.7 m horizontally. It will only travel another few meters before hitting the ground, about 1/4 second later. BobP's quadratic solution will give the exact value
so the quadratic solution is like this 4.9t^2+9t+2.2=0?
Isn't there a negative sign before the 4.9?
okay so t=2.06?
so x=22.08?
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