Posted by House on .
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,
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,
17.2ft?