Projectile motion apple throw question

[sniper_08]

Homework Statement

you have fallen into a hole in the ground. your friend throws an apple to you to keep you happy. you catch the apple 12.0 m from where your friend threw it. (basically range is 12 m). before throwing the apple to you, she threw it at the same speed directly up in the air, and it took 1.6 s to return to her had. how far below your friend are you?

since you guys can't see the diagram:
the angle above the horizontal is 30.
t = 1.6s
range or horizontal distance is 12 m
how do i find the height ( how many meters am i below the person on the cliff)
i don't know if i am doing it right how do find the height?

Homework Equations

y = (Viy)(t) + 0.5(ay)(t)^2

range = Vix x t

The Attempt at a Solution

Vix = 12 m / 1.6s = 7.5 m/s

Vix = cos30 x Vi
Vi = 7.5m/s / cos30 = 8.66 m/s

then Viy = sin 30 x 8.66m/s = 4.33 m/s

 

[gneill]

Why are you assuming that the time for your friend to toss the apple up and catch it is the same as the time it takes the apple to reach you? One velocity was directed straight up, the other is launched at an angle.

 

[sniper_08]

i don't undrstand?
ty = tx

 

[gneill]

It's your equation!

Vix = 12 m / 1.6s = 7.5 m/s

The 1.6 seconds was the time for your friend to toss the apple straight up and catch it on its return. What makes you think it applies to her tossing it down to you?

 

[sniper_08]

so how would you solve it?

 

[gneill]

I would first determine the speed at which she throws the apple (both times). The problem statement says that they are the same. So how would you go about determining that initial speed from her first toss?