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=t|v|cos(x)

x=t(42)cos35

x=42tcos35

y=t|v|sin(x)-1/2gt^2+h

y=t(42)sin35-1/2(32)t^2+0.5

y=42tsin35-16t^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=t|v|cos(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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