Question

This is physics, really. (sorry for posting here)
Do I have a brain injury?

Answer 1

Explain your condition, I'll help.

Answer 2

This relates to projectile motion range.

REMINDER:
Projectile path is symmetric about the vertex, and vertex is the maximum.

So, I know that
\(\color{#000000 }{ \displaystyle v_{y}(t)=v_{iy} -gt}\)
(velocity in the y-direction)

To find an expression for vertical displacement/position at time t, I integrate both sides,

\(\color{#000000 }{ \displaystyle \int v_{y}(t)~dt=\int~v_{iy} -gt~dt}\)

\(\color{#000000 }{ \displaystyle y(t)=v_{iy}t -\frac{1}{2}gt^2+y_0}\)

(In this context, +C is the initial vertical position from where the particle starts to travel)
~~~~~~~~~~~~~~~~~~~~~~~~~~~
Then, I know that the time to reach the top
(recall that vertical velocity at the top is zero),

\(\color{#000000 }{ \displaystyle 0=v_{iy} -gt}\)
\(\color{#000000 }{ \displaystyle gt=v_{iy} }\)
\(\color{#000000 }{ \displaystyle t=v_{iy}/g }\) (time to get to the top)

Or, in terms of the initial angle and initial speed,

\(\color{#000000 }{ \displaystyle y(t)=v_{i}\sin(\theta)t -\frac{1}{2}gt^2+y_0}\)
(the vertical position as a function of time)

\(\color{#000000 }{ \displaystyle t_{\rm ~top}=v_{i}\sin(\theta)/g }\)
(time to reach the top)

So the maximum y, (the range) is,

\(\color{#000000 }{ \displaystyle {\rm Range}=y(v_{i}\sin(\theta))}\)

And for that I get,
\(\color{#000000 }{ \displaystyle {\rm Range}=v_{i}\sin(\theta)\left(\frac{v_{i}\sin(\theta)}{g}\right) -\frac{1}{2}g\left(\frac{v_{i}\sin(\theta)}{g}\right)^2+y_0}\)

\(\color{#000000 }{ \displaystyle {\rm Range}=\frac{v_{i}^{~2}\sin^2(\theta)}{g} -\frac{1}{2}\frac{v_{i}^{~2}\sin^2(\theta)}{g}+y_0}\)

\(\color{#000000 }{ \displaystyle {\rm Range}=\frac{1}{2}\frac{v_{i}^{~2}\sin^2(\theta)}{g}+y_0}\)

\(\color{#000000 }{ \displaystyle {\rm Range}=\frac{v_{i}^{~2}\sin^2(\theta)}{2g}+y_0}\)

Answer 3

But, the formula says

\(\color{#000000 }{ \displaystyle {\rm Range}=\frac{v_{i}^{~2}\sin (2\theta)}{g}}\)

(well the initial vertical positon \(y_0\) is not that important)

Answer 4

Shouldn't \(t_{top}=\dfrac{v_i\sin(\theta)}{2g}\)?

Answer 5

No, there is \(2\theta\) there (in the formula).

Answer 6

and the initial speed is squared.

Answer 7

Oh, that formula is for horizontal range, and mine is vertical

Answer 8

I just need to verify if my result is indeed the correct maximum height...

Answer 9

Oh, nvm I got it correctly... I seee the formula online ;)

Answer 10

I guess there isn't a question, sorry for this waste. Should have webbed carefully before

Answer 11

In response to your initial question, it appears the answer is "no".

Answer 12

Oh, lol, tnx:)

Answer 13

I'll lear a lesson. Search Search and Search again, then ask. (I just have that connotation that Range is something vertical)

Answer 14

I just checked my notes. Yes the answer is correct and the range is for the horizontal range not the maximum vertical height.

Answer 15

Yes:)

Answer 16

Time to close:)

Answer 17

tnx

Answer 18

I will just finish the horizontal tange derivation for myself to leave with a good mind.

Answer 19

\(\color{#000000 }{ \displaystyle v_x(t)=v_{i}\cos(\theta) }\)

\(\color{#000000 }{ \displaystyle \int v_x(t)~dt=\int v_{i}\cos(\theta)~dt }\)

\(\color{#000000 }{ \displaystyle x(t)=v_{i}\cos(\theta)t}\)

(wil omit the constant, or the initial position in the x, since we always choose \(x_0=0\), by projectile motion. at least, I don't recal when this wasn't the case)

time to reach the top is (previously derived)
\(\color{#000000 }{ \displaystyle t_{\rm top}=\frac{v_{i}\sin(\theta)}{g}}\)

projectile path is symmetric about vertex/top, so

\(\color{#000000 }{ \displaystyle t_{\rm ~flight}=t_{\rm ~top}=\frac{2v_{i}\sin(\theta)}{g}}\)

Horizontal range,

\(\color{#000000 }{ \displaystyle {\rm R_x}=x(\frac{2v_{i}\sin(\theta)t}{g})=v_{i}\cos(\theta)\frac{2v_{i}\sin(\theta)t}{g}}\)

\(\color{#000000 }{ \displaystyle {\rm R_x}=\frac{2v_{i}^{~2}\sin(\theta)\cos(\theta)}{g}=\frac{v_{i}^{~2}\sin(2\theta)}{g}}\)

Answer 20

boom boom :)