Derive a general expression for range when launch angle=0

[aron silvester]

1. The problem statement, all variables, and given/known data

Homework Equations

The Attempt at a Solution

So basically I took vertical direction formula and I solved for "t". I then took the horizontal direction formula and solved for "x" or we can also call x the "Range" since that's what the problem is asking for. Is this right?

 

[TSny]

Correct. Good job!

 

[aron silvester]

Correct. Good job!

Hey thanks, but my professor got this. The equation that I got is the same as the expression on the left side of the addition sign, and then he's adding an equation inside a square root. Why is that?

 

[CWatters]

I haven't looked at your professors solution but I think yours is wrong.

If the launch angle is zero then sin(θ) = 0 and the range will always be zero unless it's fired from some initial height y₀. Yet you state y₀=0.

The problem statement mentions "h" which I think meant to be the initial launch height eg it's fired from the top of a hill of height h.

 

[robax25]

i mean if you set up theta=0, you will not get the formula. then you get that formula, R=V0*t
= vo*√(2h/g)

 

[TSny]

Hey thanks, but my professor got this. The equation that I got is the same as the expression on the left side of the addition sign, and then he's adding an equation inside a square root. Why is that?
View attachment 212478

As @CWatters noted, you solved the problem for firing the projectile with a nonzero angle over level ground, so the projectile returns to the same height as it was fired from. But it looks like you were meant to solve a different problem where the projectile is fired horizontally ($\theta = 0$) off a cliff of height h. I apologize for not catching that earlier.

 

[aron silvester]

I haven't looked at your professors solution but I think yours is wrong.

If the launch angle is zero then sin(θ) = 0 and the range will always be zero unless it's fired from some initial height y₀. Yet you state y₀=0.

The problem statement mentions "h" which I think meant to be the initial launch height eg it's fired from the top of a hill of height h.

As @CWatters noted, you solved the problem for firing the projectile with a nonzero angle over level ground, so the projectile returns to the same height as it was fired from. But it looks like you were meant to solve a different problem where the projectile is fired horizontally ($\theta = 0$) off a cliff of height h. I apologize for not catching that earlier.

I finally solved it, but I have a question. When I used the quadratic formula, why did I have to multiply the inside of the square root by (v^2cos^2(θ))/g^2 and again by (v^2cos^2(θ))/1 ?? The part where I do this is in pink ink.

 

[TSny]

I finally solved it, but I have a question. When I used the quadratic formula, why did I have to multiply the inside of the square root by (v^2cos^2(θ))/g^2 and again by (v^2cos^2(θ))/1 ?? The part where I do this is in pink ink.
View attachment 212534

When you bring something into the square root, you have to square it. The something here is $\frac{v_0^2 \cos ^ 2\theta}{g}$. So, after bringing it into the square root, the $v_0$ and $\cos{\theta}$ are raised to the 4th power.