Projectile Motion in Elevator: Peach Pit Launch Homework

[halo9909]

Homework Statement

A rude tourist throws a peach pit horizontally with a 7.0 m/s velocity out of an elevator cage.
(a) If the elevator is not moving, how long will the pit take to reach the ground, 11.0 m below?
s
(b) How far (horizontally) from the elevator will the pit land?
m
(c) He throws the next pit when the elevator is at the same height but moving upward at a constant 9.5 m/s velocity. How long will it take this pit to land?
s
(d) How far away will this pit land?

Homework Equations

X Y
d h
v vi
t vf
t
a

d=.5at^2
basic trigonometry

The Attempt at a Solution

I am unsure how to approach this, since I don't understand this, what would need tobe done to find time in part1

 

[Hootenanny]

Start my splitting the motion into two components, vertical and horizontal motion. Now, considering the vertical motion only, how long will it take the peach to hit the ground?

 

[halo9909]

yeah i got part a and b
t=Squareroot(2d/9.8)

and got the "t" to be 1.498sec and multplied the time by 7 and got "b" which is 10.483m
but for the other two it is left me confused, on how to use like forumulas

 

[Hootenanny]

In the previous case, the initial velocity only had an horizontal component, but in the current case the initial velocity has both a horizontal and vertical component. However, the approach is much the same.

Start by splitting the motion into the horizontal components. Which kinematic formula do you think would be appropriate for part (c)?