Projectile motion Q3 (A)

[Aaron Curran]

Here is the problem (Q3 (A));

Here's what I have so far;

Any help would be appreciated, thank you!

 

[RUber]

It appears you made an algebra error when you went from $21 \cos \alpha = \frac {30}{ t}$ to $\cos \alpha = \frac {630}{ t}$.
Also, I would first find your time of impact based on your horizontal component.
With that time of impact, you can make an expression that is in terms of alpha only.

 

[Aaron Curran]

It appears you made an algebra error when you went from $21 \cos \alpha = \frac {30}{ t}$ to $\cos \alpha = \frac {630}{ t}$.
Also, I would first find your time of impact based on your horizontal component.
With that time of impact, you can make an expression that is in terms of alpha only.

Yes I realized this after I posted, I'll try that, thanks!

 

[RUber]

I made a substition of x = cos^2(alpha) and manipulated it into a quadratic equation to get the two solutions.
part b comes right out of part a if you have already solved for time of impact.

 

[TSny]

Aaron,

Something else to note: If you have the expression $\tan \alpha = \frac{A}{B}$, then you cannot conclude that $\sin \alpha = A$ and $\cos \alpha = B$.

 

[Aaron Curran]

I made a substition of x = cos^2(alpha) and manipulated it into a quadratic equation to get the two solutions.
part b comes right out of part a if you have already solved for time of impact.

Could you elaborate on this a bit? Not quite sure what you mean..

 

[RUber]

Do you have an expression that can be written as some trig functions of alpha = 0?
Use sin = sqrt(1-cos^2) to convert everything in terms of cos.