Shooting in a π/4 Angle: Is it Possible?

[fer Mnaj]

Homework Statement

Proof that in a parabolic motion with r0 = 0 and v_y0 = 0, the maximum horizontal range (in the x direction) is given by shooting with an angle of π / 4 inclination about the horizontal direction.

Relevant Equations

y(t)= y_0 +V_0y t-1/2 gt^2
x(t)= x_0+V_0x t

So the statement says that in r(0)=0, so it departs from the origin. And it also says that v_y0 = 0, thus meaning thar it´´´ s position in the y direction is Constant. Is shooting in a π / 4 angle still plausible? or there´ s no way that can happen? So far I´ ve thought since there´ s no acceleration affecting the movement, its position must be determine by x(t)= x_0+V_0x t which is x(t)= x_0+V_0cos (θ) t.
So what do you think?

 

[haruspex]

Homework Statement:: Proof that in a parabolic motion with r0 = 0 and v_y0 = 0, the maximum horizontal range (in the x direction) is given by shooting with an angle of π / 4 inclination about the horizontal direction.

Is shooting in a π / 4 angle still plausible?

It does seem nonsensical.
Are you sure you have quoted the question exactly? Presumably it does not quite have "v_y0=0", but perhaps $v_y(0) =0$, yes?
Is it a translation?

 

[etotheipi]

And it also says that v_y0 = 0, thus meaning thar it´´´ s position in the y direction is Constant.

What if you stand at the top of a cliff, and throw the projectile horizontally out toward the sea? The initial vertical component of velocity would be zero, but it's position in the $y$ direction would definitely not b constant...

In any case, I have no idea how that would relate to proving that the optimum angle above the horizontal is $\pi/4$.

I think you will need to upload the original problem statement

 

[jbriggs444]

What if you stand at the top of a cliff, and throw the projectile horizontally out toward the sea? The initial vertical component of velocity would be zero, but it's position in the $y$ direction would definitely not b constant...

In any case, I have no idea how that would relate to proving that the optimum angle above the horizontal is $\pi/4$.

In the case of launching a projectile from the top of a cliff, the optimum launch angle to maximize horizontal range onto the flat plain below will not be $\frac{\pi}{4}$.

In the case of a projectile launched with a vertical velocity component of zero and an angle of $\frac{\pi}{4}$ above the horizontal, the horizontal component will be zero as well. That projectile will drop like a stone. The maximum range will be equal to the minimum range which will be zero. This will be achieved by all launch angles, including $\frac{\pi}{4}$.

 

[etotheipi]

@jbriggs444 yes there is something that OP has omitted or transcribed incorrectly, because as stated the OP makes no sense

 

[kuruman]

If, instead of $v_{y0}=0$, the initial condition were $y_0=0$, i.e. the projectile lands at its original height, the whole thing makes sense. However, for the title question to make more sense if it were "Is the maximum horizontal range still given by π / 4 in parabollic motion when the projectile launcher moves relative to the ground?" Something lost in translation?

 

[haruspex]

if it were "Is the maximum horizontal range still given by π / 4 in parabollic motion when the projectile launcher moves relative to the ground?"

I.e., it should say "if $v_x(0)>0$"? That would make sense, but it's two typos.

 

[etotheipi]

I know that Einstein is rumoured to have said something along the lines of "if I had one hour to solve a problem, I'd spend 55 minutes thinking about the problem and 5 minutes thinking about solutions"... but I think we are taking this a little bit too literally

It seems like OP has done an Elvis Presley on us...