Find the max height of the second projectile relative to seal level

[PhysicsFTW22]

Homework Statement

A projectile is launched at 110.0 m/s at 40.0° above the horizontal off a cliff that is 250.0 m above sea level. After 4.00 s of flight, this projectile launches a second projectile at 50.0 m/s directed at 15.0° below the horizontal.
a) Find the max height of the second projectile relative to seal level.
b) Find the range of the second projectile when it reaches sea level.

Homework Equations

In x-direction: x=(vicos∅)t , vi=vf , vix= vicos∅
In y-direction: y=yi + vit + .5at^2 , v=vi+at , vi=visin∅

The Attempt at a Solution

I got the answer to part a, 472m. However, I am lost at trying to find an answer to part b. Using x=(vi)(t)=(110cos40)(4)=337m, I used that as the range the first projectile goes before releasing the second projectile. Now in trying to find how far the second projectile goes when it hits sea level, I plan on adding that to the first range. Any thoughts?

 

[danielu13]

Using position functions:

$x=v$_i$*t*cosθ$

$y=v$_i$*t*sinθ-\frac{1}{2}*9.81*t^2+y$_o

You get the position of projectile 1 as (337.060, 454.347) when the second projectile is released. Using velocity equation:

$v=v$_i$+at$

The velocity at t(4) is (84.265 $\hat{i}$, 31.467 $\hat{j}$), neglecting drag.

Projectile 2 is released at 50 m/s at 15°, which is added to the velocity of projectile 1, resulting in an initial velocity of (132.561 $\hat{i}$, 18.526 $\hat{j}$). Do note that the launch velocity in the y direction was subtracted from the value for projectile 1 because it was in the downward direction, and I believe this was at a time that the projectile was still rising; if it was falling you would just add the initial velocity of projectile 2 relative to projectile 1, which was 12.941 $\hat{j}$, to the velocity of projectile 1 at t(4).

You then use this velocity to find the time at which projectile 2 will hit the ground, then use the position function to find the range. I know this was a bit of a mess, but it should help. I can work it out further if necessary.