Ball falls from top of cliff; kinematics

[dansunefusee]

Homework Statement

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/s²
H (height of cliff)= -17.4m
V1 (final velocity)= ?

Ball 2:
u2= V1= ?

Unknowns:
t= ?
x (where balls meet above the ground)= ?

Homework Equations

V²=u² + 2ax
x = u(t) + 0.5(g)(t)²

The Attempt at a Solution

First, I calculated the final velocity (V1) of Ball 1 as so:
V1²=u² + 2ax
V1= sqrt[(0m/s)² + 2(-9.81m/s²)(-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!

 

[rl.bhat]

Hi dansunefusee. welcome to PF.

Both the balls start simultaneously. When they meet they must have traveled same duration t s. If x is the distance from the top of the cliff, then for the falling ball
-x = -1/2*g*t^2...(1)
For the ball going up
h - x = ut - 1/2*g*t^2 ...(2)
Where h is the height of the cliff. Solve the two equations and find the time t.

 

[dansunefusee]

Thank you! :) I'll need all the help I can get from PF this year. Haha.

To make sure I am doing this right, I would need to combine the 2 equations just to find time?

Given equations:
1) -x = -1/2*g*t^2
2) h - x = ut - 1/2*g*t^2

Therefore, I would plug the 1st equation for the -x variable in the 2nd equation to get:

h - (-1/2*g*t²) = ut - 1/2*g*t²

Then isolate the time (t) variable by itself?

 

[kalupahana]

Hey , after you take the time(t) , substitute it to (1) equation , then x will come,

 

[rl.bhat]

Thank you! :) I'll need all the help I can get from PF this year. Haha.

To make sure I am doing this right, I would need to combine the 2 equations just to find time?

Given equations:
1) -x = -1/2*g*t^2
2) h - x = ut - 1/2*g*t^2

Therefore, I would plug the 1st equation for the -x variable in the 2nd equation to get:

h - (-1/2*g*t²) = ut - 1/2*g*t²

Then isolate the time (t) variable by itself?

This step is wrong.

It should be h - 1/2*g*t² = ut - 1/2*g*t²