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

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 button-pushing

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.

1. 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.

posted by Steve
2. which of the following is a highest value ?15/16 b. 5/8 c. 3/4.

posted by kaka

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