Math(Please answer)
posted by Hannah on .
This is the same equation as before just with different numbers.
y=yo + (vo sin Q) t  1/2gt^2
This time the numbers are:
0 = 1.005 + (3.021 sin 30)t  1/2(9.8)t^2
4.9t^2  1.5105  1.005 = 0
This is the part where I am stuck. Did I do this correctly? sin 30 = 0.50 X 3.021 = 1.5105 and then I would subtract this from 1.005
so then it would be 1.5105  1.005 = 0.5055 / 4.9 and square root it.
Math(Please check)  Reiny, Thursday, September 13, 2012 at 10:29am
You are making a similar mistake to yesterday's error.
Yesterday you just magically tagged a t at the end of the first term, now you are dropping the t from the second term
your equation of
0 = 1.005 + (3.021 sin 30)t  1/2(9.8)t^2
looks ok, assuming your replacement values were correct
then
0 = 1.005 + 1.5105t  4.9t^2 , since sin 30‹ = 1/2
4.9t^2  1.5105t  1.005 = 0 , look at yours
you will have to use the quadratic formula
t = (1.5105 } ã(1.5105^2  4(4.9)(1.005) )/9.8
= .... you do the buttonpushing
Math(Please check)  Steve, Thursday, September 13, 2012 at 10:33am
I think you dropped a t:
4.9t^2  1.5105t  1.005 = 0
the coefficients are correct, so now you just have to solve the quadratic equation.
t = 0.154 } ã0.229
How did you get 0.154 and 0.229? When I try to complete the quadratic equation I am not getting these numbers.

use the quadratic formula.
t = [1.5105 +/ sqrt(1.5105^2 + 4*4.9*1.005)]/(2*4.9)
= [1.5105 +/ sqrt(21.9796)]/9.8
= (1.5105 +/ 4.6882)/9.8
= 0.324 or 0.632
Hmm. you are correct. Don't know how I came up with those bogus values. Probably cut/pasted from somewhere else. Good catch. 
which of the following is a highest value ?15/16 b. 5/8 c. 3/4.