Posted by . on Saturday, September 18, 2010 at 10:25am.
A shell is launched at 140 m/s 56.0 degrees above the horizontal. When it has reached its highest point, it launches a projectile at a velocity of 110 m/s 35.0 degrees above the horizontal relative to the shell.
1) Find the maximum height (in m) about the ground that the projectile reaches.
2) Find its distance (in m) from the place where the shell was fired to its landing place when it eventually falls back to the ground.
Solution/explanation, please? Thanks!

Physics  bobpursley, Saturday, September 18, 2010 at 10:36am
first, figure the height of the shell.
vertical velocity at launching is 140sin56.
At the top, the vertical velocity of the shell is zero.
Vf^2=Vi^22*g*height solve for height.
Now, the second problem, the second projectile. Do it the same way, except vi now is 110sin35
add the two heights.
Now for distance. I will leave that to you, find the distance to the first max height (vcostheta*time), and then the second distance. add them. Careful on your calculations. 
Physics  Anonymous, Thursday, October 25, 2012 at 11:15pm
jgm