How Do Two Balls Meet Mid-Air When Thrown at Different Times and Heights?

[Arshad_Physic]

Homework Statement

A ball is thrown with an upward velocity of 5 m/s from the top of a 10-m high building. One second later, another ball is thrown vertically from the ground with a velocity of 10m/s. Determine the height from the ground where the two balls bass each other.

Homework Equations

1) v=v_ot + at
2) s=s_o + v_ot + 0.5at²
3) v² = v_o² + 2a(s-s_o)

The Attempt at a Solution

I know that this is a basic physics problem. I have taken University Physics 1 before, and made an A, but I am stuck in this problem lol.

I am assuming that we do something like y₁=y₂ and then just solve the equation, get time. And then plug the time in. Or we can do t₁=t₂

But I am having problem making the equations.

I did this:

y₁=v_o1t₁ + 0.5at₁²

V^o1 = 5 m/s
a = 9.8 m/s²

y₂=v_o2t₂ + 0.5at₂²

t₁ = t₂ - 1

v_o2 = 10 m/s

The problem is, my t₁ is coming out to be -11/7 Lol :)

Please help!

Thanks,
Arshad

 

[jahrollins]

y1 doesn't equal y2. The ball is being thrown from the building so their displacements will be different.