How Do You Solve for Time in Projectile Motion Equations?

[Zashmar]

15=35 sin Theta*t -4.9t^2

I have tried solving for t, I do not know how to...

please help

thank you

 

[dauto]

What is Θ? (If you click on the "Go Advanced" button you will be able to put in the Greek letters as needed it also allows you to use superscripts.

You did not use parenthesis so I can tell you mean
15=35 sin (Θ)*t -4.9t², or
15=35 sin (Θ*t) -4.9t², or
15=35 sin (Θ*t -4.9)t², or
15=35 sin (Θ*t -4.9t)², or
15=35 sin (Θ*t -4.9t²), or

The answer to your question depends on that. Bottom line: You need to learn to use parenthesis as needed.

 

[Zashmar]

15=35 sin (Θ)*t -4.9t^2

Sorry,

 

[Chestermiller]

What is the solution to the equation at² + bt + c = 0 ?

 

[Zashmar]

((-b(+-)sqrt(b^2 -4ac))/2a

 

[Zashmar]

I know the quadratic formula, but how will I do it if j have theta as well?

 

[mafagafo]

b = 35 * Sin(Θ) in this case. It is a symbolic formula, if it is a quadratic, it is valid.
Just make your equation "fit" the format of the formula and you will see what is a, what is b and what is c.
Then solve it.

 

[Zashmar]

So...

x= $(4.9 \pm \sqrt{24.01+2100 sin\vartheta}/(70 sin \vartheta)$

 

[Chestermiller]

So...

x= $(4.9 \pm \sqrt{24.01+2100 sin\vartheta}/(70 sin \vartheta)$

It doesn't look like you applied the formula correctly. What did you use for a, b, and c?

Chet

 

[Zashmar]

15=35 sin (Θ)*t -4.9t2

a=35 sinΘ
b=-4.9
c=-15

 

[mafagafo]

This is wrong. You posted the correct formula above but identified wrong values for a, b and c.
Also, you missed a parenthesis.

 

[mafagafo]

a is the coefficient of the t^2 term, b is the coefficient of the t term and c is the independent term. Try again. Why don't you rearrange it as I suggested, it is a good practice if a, b and c don't seem ridiculously obvious.

 

[Zashmar]

Okay

So I have it now, how would i simplify the following (I have subbed it into x=35cos$\vartheta*t$)?

x=35cos$\vartheta$($-35 sin\vartheta \pm \sqrt{(35sin \vartheta)^2 -294}/-9.8$

I tried to put in the parentheses but it would not let me

 

[Chestermiller]

So I have it now, how would i simplify the following (I have subbed it into x=35cos$\vartheta*t$)?

x=35cos$\vartheta$($-35 sin\vartheta \pm \sqrt{(35sin \vartheta)^2 -294}/-9.8$

I tried to put in the parentheses but it would not let me

You need to get the algebra right before we can help you simplify anything.

Chet

 

[Zashmar]

all terms except 35cos theta, should be divided by -9.8. I tired to show the parentheses but it wouldn't work

 

[mafagafo]

I may be missing something, but why is there a cos(theta) anyways?

 

[Zashmar]

Im subbing it into another equation

 

[mafagafo]

Be clearer, are you trying to do:
1)$cos{(\theta \cdot t)}$
or
2)$cos{(\theta)} \cdot t$

Also, it seems to me that your solution for x is still not correct. Check your signs.

 

[Zashmar]

The second option :)

 

[mafagafo]

You need to get the algebra right before we can help you simplify anything.

As he said, redo it. From what I can see, your signs are wrong.

 

[dauto]

It seems you're solving a problem of projectile motion. Is this correct? Would you post the problem so we can help you more effectively?