(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A ball is thrown upward from the edge of a 20m building with velocity V. One second later a ball is dropped from the same building. With what velocity must the first ball be thrown so that both balls land at the same time.

2. Relevant equations

x(t)=x+vx+1/2(ax)^2

I'm am using x1=the first ball

I'm using x2=the second ball

vx^2=2(ax)d^2

3. The attempt at a solution

So I solved for how long it will take x2 to fall the height of the building with the equation

0=20-1/2(10m/s)*t^2

so t=2s

So, I guess since it takes 2 seconds for x2 to fall the distance of the building, plus the extra second, x1 must take 3 seconds to fall.

So know I guess I need to find with what velocity the first ball must be thrown so that it takes three seconds to hit the ground.

The way the professor explained in lecture was by using the equation that velocity squared is equal to twice the acceleration times the distance.

so Vx^2=2(ax)d

he also divided the times into the time it takes from the ball to go from the roof to the peak of the throw, then from the peak to the roof, and then from the roof to the ground.

So I guess the time from the roof to it's maximum height is by the distance equation also. So for x1

y(t)=vx(t)+1/2(-10m/s)t^2

y(t)=vx(t)-5t^2

and since vx^2=2(ax)d

vx^2=2(-10)d

vx^2=-20d

I dunno, I guess I didn't really follow how my professor did this. Also, i'm having a hard time realizing what the difference between distance and position is. I know that if I can find the velocity upward and the velocity as the ball is passing the roof on the way down. I can find the time for the final distance down the building using that initial velocity. I dunno, I'm lost. Any help is appreciated!

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Constant Acceleration Problem

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