I am off by one decimal and I cannot see the error in my work

[pemby]

Here is a screenshot of the question I deleted on stack physics after they directed me to this forum.

Eratta:
FYI I have tried with g = -9.80 and 58.9 m when the correct solution (I think) is 5.9 rounded two significant figures. Thanks for any help!

 

[ehild]

I think you used -(-9.8) for g in the height formula. The formula for the displacement in case of uniformly accelerating motion is y=v_ot+a/2 t², and a= - 9.8 now. Also, the second term on the right side should be vtan(30°).

 

[pemby]

Hi, I am attempting to derive a solution based on equations on the 1st page of this document.
http://themcclungs.net/physics/download/H/2_D_Motion/Projectile Cliff.pdf

" The formula for the displacement in case of uniformly accelerating motion is y=vot+a/2 t2"
In the spirit of understanding where formulas come from could you derive that from the equations I am using to derive my solution?

I

 

[Chestermiller]

I think you used -(-9.8) for g in the height formula. The formula for the displacement in case of uniformly accelerating motion is y=v_ot+a/2 t², and a= - 9.8 now. Also, the second term on the right side should be vtan(30°).

I agree with your g assessment, but think you should reconsider that advice about the tan 30 degrees vs sin 30 degrees.

 

[ehild]

I agree with your g assessment, but think you should reconsider that advice about the tan 30 degrees vs sin 30 degrees.

You are right, it is tangent in the formula y(x).

 

[ehild]

Hi, I am attempting to derive a solution based on equations on the 1st page of this document.
http://themcclungs.net/physics/download/H/2_D_Motion/Projectile Cliff.pdf

The text you refer to would use +9.8 m/s² for the acceleration due to gravity.
" The formula for the displacement in case of uniformly accelerating motion is y=volt+a/2 t2"
In the spirit of understanding where formulas come from could you derive that from the equations I am using to derive my solution?

I[/QUOTE]
Your solution is wrong. for g, you have to use the positive value, g=9.8 m/s².

 

[pemby]

Hi all...

Tangent = sin/cos = sin*1/cos that is just rewriting the expression. It does not change the value.

$${ y }_{ f }\quad =\quad -\frac { 1 }{ 2 } g{ \frac { x }{ \left( v\cos { \left( \theta \right) } \right) } }^{ 2 }+\left( v\sin { \left( \theta \right) } \right) \frac { x }{ \left( v\cos { \left( \theta \right) } \right) } \quad +\quad { y }_{ 1 }\quad =-\frac { 1 }{ 2 } g{ \frac { x }{ \left( v\cos { \left( \theta \right) } \right) } }^{ 2 }+\frac { v }{ v } \frac { \sin { \left( \theta \right) } }{ \left( \cos { \left( \theta \right) } \right) } x+y1\quad =-\frac { 1 }{ 2 } g{ \frac { x }{ \left( v\cos { \left( \theta \right) } \right) } }^{ 2 }+\tan { \left( \theta \right) } x+y1$$

http://www.wolframalpha.com/input/?i=-1/2(-9.81)*(x/(v*cos(30)))^2+v*sin(30)*x/(v*cos(30))+y1=0,+x=16.5,+v=6+solve+for+y1

http://www.wolframalpha.com/input/?i=-1/2(-9.81)*(x/(v*cos(30)))^2+sin(30)/(cos(30))*x+y1=0,+x=16.5,+v=6+solve+for+y1

http://www.wolframalpha.com/input/?i=-1/2(-9.81)*(x/(v*cos(30)))^2+tan(30)*x+y1=0,+x=16.5,+v=6+solve+for+y1

 

[vela]

Hi, I am attempting to derive a solution based on equations on the 1st page of this document.
http://themcclungs.net/physics/download/H/2_D_Motion/Projectile Cliff.pdf
"The formula for the displacement in case of uniformly accelerating motion is y=volt+a/2 t2"

In the spirit of understanding where formulas come from, could you derive that from the equations I am using to derive my solution?

If you start with the assumption that acceleration $a_y$ is constant, you can derive the equation $\Delta y = v_{0_y} \Delta t + \frac 12 a_y \Delta t^2$. In the case of free fall, following the usual sign conventions, you have $a_y = -g$ where the minus sign denotes the direction of the acceleration and $g=+9.80~\rm{m/s^2}$ is the magnitude of the acceleration due to gravity.