you would get
4.9t^2 - 1.005t = 0
t(4.9t - 1.005) = 0
t = 0 or t = 1.005/4.9 = 201/980 or appr .205
Using the equation
y=yo + (vo sin Q) t - 1/2gt^2 I have to find t. The values for y, yo and vo were given to me.
So I set the equation up like this:
0=1.005 + (3.107 sin 0) t - 1/2(9.8)(t^2
The sin of 0 is 0 times 3.107 is 0.
So 0 = 1.005 t - 4.9 t^2
Now I am stuck. I am not sure how to solve this part. Thank you for your help!
8 answers
Where do you get the 1.005t? It appears to me that you just stuck a t onto the number 1.005.
The 1.005 was the value given for yo, the angle given was 0 and vo was 3.107. The t I just took from the equation
y=yo + (vo sin Q) t -1/2 gt^2.
I need to find t using this equation but I wasn't sure how to do it.
y=yo + (vo sin Q) t -1/2 gt^2.
I need to find t using this equation but I wasn't sure how to do it.
I never even looked at the equation that closely, just assumed she did the subbing correctly
now it would simply be
4.9t^2 = 1.005
t = ± √(1.005/4.9) = appr .45
now it would simply be
4.9t^2 = 1.005
t = ± √(1.005/4.9) = appr .45
I may be dense but I still don't get it. if yo is 1.005, that's the number that goes into the equation for yo. There is no t there. The equation would have to be yo*t for that to be true. right?
Ahh! I haven't forgotten ALL my math. Just most of it
assuming her values were valid and she subbed them in correctly
0=1.005 + (3.107 sin 0) t - 1/2(9.8)(t^2
4.9t^2 - 0 - 1.005 = 0
4.9t^2 = 1.005
t^2 = 1.005/4.9
t = ± √(1.005/4.9) = appr .45
0=1.005 + (3.107 sin 0) t - 1/2(9.8)(t^2
4.9t^2 - 0 - 1.005 = 0
4.9t^2 = 1.005
t^2 = 1.005/4.9
t = ± √(1.005/4.9) = appr .45
Ok so just to make sure I take the square root of 1.005/4.9= 0.45 which would be both positive and negative but I only take the negative correct? Thank you for your help. This makes sense now because I have to compare this value to the actual which was 0.44 so I am not to far off.