2D Projectile Motion problem, is there a better solution?

[Habeebe]

Homework Statement

Romeo is chucking pebbles gently up to Juliet's window and he wants the pebbles to hit the window with only a horizontal component of velocity. He is standing at the edge of a rose garden 8.0 m below her window and 9.0 m from the base of the wall. How fast are the pebbles going when they hit her window?

Homework Equations

Is this asking the equations I use or equations provided? I guess the question that I'm answering, since I'm confident in my solution, is if there exists an easier route to the solution.

The Attempt at a Solution

All units are in meters, seconds, and degrees. Since I don't convert between I won't put them in for ease of typing.

Since I know the trajectory starts at (0,0) and has a vertex at (9,8), I found the equation of the parabola.

$$y=a(x-9.0)^2+8.0$$
$$a|_{(0,0)}=\frac{-8}{81}$$ substitute a back in and expand
$$y=\frac{-8x^2}{81}+\frac{16x}{9}$$ differentiate
$$y'=\frac{-16x}{81}+\frac{16}{9}$$ plugged in 0 for x to find the slope at origin and got 16/9 then found what angle that represents
$$\theta=arctan(\frac{16}{9})=60.6 degrees$$ Used V^2 under constant acceleration formula
$$V_y^2=V_{y0}^2-19.6*\Delta y$$
$$0=V_{y0}^2-19.6*8$$
$$V_{y0}=12.52$$ I know the vertical component and angle, so I find the horizontal component
$$V_{x0}=\frac{12.52}{tan(60.6)}=7.06 m/s$$

Since the horizontal velocity doesn't change, and the vertical velocity is zero, then the velocity it hits the window is 7.06, the same as the starting horizontal velocity.I feel like the way I got here was really cumbersome. Was there an easier way for me to find the same thing?

 

[gneill]

Homework Statement

Romeo is chucking pebbles gently up to Juliet's window and he wants the pebbles to hit the window with only a horizontal component of velocity. He is standing at the edge of a rose garden 8.0 m below her window and 9.0 m from the base of the wall. How fast are the pebbles going when they hit her window?

Homework Equations

Is this asking the equations I use or equations provided? I guess the question that I'm answering, since I'm confident in my solution, is if there exists an easier route to the solution.

The Attempt at a Solution

All units are in meters, seconds, and degrees. Since I don't convert between I won't put them in for ease of typing.

Since I know the trajectory starts at (0,0) and has a vertex at (9,8), I found the equation of the parabola.

$$y=a(x-9.0)^2+8.0$$
$$a|_{(0,0)}=\frac{-8}{81}$$ substitute a back in and expand
$$y=\frac{-8x^2}{81}+\frac{16x}{9}$$ differentiate
$$y'=\frac{-16x}{81}+\frac{16}{9}$$ plugged in 0 for x to find the slope at origin and got 16/9 then found what angle that represents
$$\theta=arctan(\frac{16}{9})=60.6 degrees$$ Used V^2 under constant acceleration formula
$$V_y^2=V_{y0}^2-19.6*\Delta y$$
$$0=V_{y0}^2-19.6*8$$
$$V_{y0}=12.52$$ I know the vertical component and angle, so I find the horizontal component
$$V_{x0}=\frac{12.52}{tan(60.6)}=7.06 m/s$$

Since the horizontal velocity doesn't change, and the vertical velocity is zero, then the velocity it hits the window is 7.06, the same as the starting horizontal velocity.I feel like the way I got here was really cumbersome. Was there an easier way for me to find the same thing?

Hi Habeebe, welcome to Physics Forums.

There are other ways to reach the result that you found. One such method would be to realize that you want the pebbles to reach the window at the apex of their trajectory arc, where the vertical velocity is zero and all that's left is the horizontal component of their initial velocity.

So, first calculate the time required for a pebble to reach the desired maximum height. This can be done using the known acceleration due to gravity for the vertical motion (g) and the desired maximum height (8.0m). Then use that time to find the x-component of the velocity given the required horizontal distance.

 

[Habeebe]

Thanks. That method took the last four lines of my paper, where the rest of the space was consumed by my method.