Posted by House on Monday, October 6, 2008 at 3:26pm.
A football is kicked with an initial velocity of 42 ft/sec at an angle of 35 degrees with the horizontal. How far has the football traveled horizontally after 0.5 sec? round to nearest tenth.
Could someone explain this please?
x=tvcos(x)
x=t(42)cos35
x=42tcos35
y=tvsin(x)1/2gt^2+h
y=t(42)sin351/2(32)t^2+0.5
y=42tsin3516t^2+0.5
0.5=42tcos35
t=0.5/42cos35
t=0.00975181
I don't know, I just followed an example in my book

Math(repost)  bobpursley, Monday, October 6, 2008 at 4:33pm
Let me edit your work:
HORIZONTAL>>
x=tvcos(x)
x=t(42)cos35
x=42tcos35
so this is the equation of the horizontal component of the football. At time 1/2 second, put that in for t, and you have the horizontal distance.
The rest of your work is not related.

Math(repost)  House, Monday, October 6, 2008 at 4:43pm
17.2ft?
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