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

A ball is dropped from rest from the top of a cliff that is 17.4 m high. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is exactly the same as that with which the first ball eventually hits the ground. In the absence of air resistance, the motions of the balls are just the reverse of each other. Determine how far below the top of the cliff the balls cross paths.

Ball 1:

u1 (initial velocity)= 0 m/s

g= -9.81 m/s^{2}

H (height of cliff)= -17.4m

V1 (final velocity)= ?

Ball 2:

u2= V1= ?

Unknowns:

t= ?

x (where balls meet above the ground)= ?

2. Relevant equations

V^{2}=u^{2}+ 2ax

x = u(t) + 0.5(g)(t)^{2}

3. The attempt at a solution

First, I calculated the final velocity (V1) of Ball 1 as so:

V1^{2}=u^{2}+ 2ax

V1= sqrt[(0m/s)^{2}+ 2(-9.81m/s^{2})(-17.4m)

V1= 18.4673m/s ~18.47m/s

Therefore, the u2 (initial velocity) of Ball 2 is also 18.47m/s.

I know that the next step is to calculate the time by using x = u(t) + 0.5(g)(t)^{2}, but I'm not sure on how to manipulate the equation to fit the criteria of this problem. Please help me!

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# Homework Help: Ball falls from top of cliff; kinematics

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