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

AI Thread Summary
To determine where two balls meet mid-air when thrown at different times and heights, one ball is thrown from a 10-meter building with an initial velocity of 5 m/s, while the second ball is thrown from the ground with a velocity of 10 m/s one second later. The equations of motion for both balls need to be set up correctly to find their heights as functions of time. The first ball's height equation considers its initial height and upward motion, while the second ball's equation starts from the ground. The key challenge is correctly relating the time variables for both balls, as they are thrown at different times. Solving the equations will yield the height at which they intersect.
Arshad_Physic
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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=vot + at
2) s=so + vot + 0.5at2
3) v2 = vo2 + 2a(s-so)

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=vo1t1 + 0.5at12

Vo1 = 5 m/s
a = 9.8 m/s2

y2=vo2t2 + 0.5at22

t1 = t2 - 1

vo2 = 10 m/s

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

Please help!

Thanks,
Arshad
 
Last edited:
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y1 doesn't equal y2. The ball is being thrown from the building so their displacements will be different.
 
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