FInd the height where two object pass each other.

First off, the acceleration should be 4.91t², not 4.95t². However, it's much better practice just to leave it as gt²/2.

Now, allow me to look at the problem in a way that makes more sense to me, and may help you better understand it.
Let's make an equation for the height of the dropping ball (ball 1):

[itex]\displaystyle h_1 = 42 - 6t - \frac{gt^2}{2}[/itex]

Now, an equation for the upwards-traveling ball (ball 2):

[itex]\displaystyle h_2 = 55t - \frac{gt^2}{2}[/itex]


The question asks for the height when [itex]\displaystyle h_1=h_2[/itex], or:

[itex]\displaystyle 42 - 6t - \frac{gt^2}{2}=55t - \frac{gt^2}{2}[/itex]

I'll let you continue from there. Hope that helps; If you have more trouble just ask.

 

The ball is thrown downward at 6 m/s. If you define the starting height to be 42 and the ground to be 0, this means the initial velocity is negative. Another way to put it is that it was thrown at 6m/s toward the ground (toward 0).

If the height were "42 + 6t - gt²/2" as you say, then that would imply that the ball was thrown upwards at 6m/s (and not downward). The height would first increase to some maximum above 42, then fall past 42 again.

 

ok


now my try....

again assume h distance from ground as meeting point at time t

so h=55t-0.5(9.8)t²(acceleration=-ve as velocity is decreasing)

and 42-h=6t+0.5(9.8)t²

solve for h

i think it gives same result as that of dude14