Question

A blue ball is thrown upward with an initial speed of 20.0 m/s, from a height of 0.7 meters above the ground. 2.4 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 9.8 m/s from a height of 22.6 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

Answer 1

And the question is...

Answer 2

I assume the question is when will the two balls "meet", i.e. at the same vertical elevation.

Answer 3

... and at what elevation will they "meet".

Answer 4

Are you familiar with the kinematics equation:
x(t)=x(0)+vi*t-(1/2)gt^2
for the vertical position of an object initially at x(0) and thrown upwards with an initial velocity of vi subject to acceleration (downwards) of gravity of g=9.81 m/s^2.

For the blue ball:
x(0)=0.7 m
vi=20 m/s
so
x(t)=0.7+20t-(9.81/2)t^2

For the red ball:
x(0)=22.6 m
vi=-9.8 m/s (downwards)
but since it is thrown 2.4 seconds later, we replace t by t-2.4, therefore
x(t)=22.6-9.8(t-2.4)-(9.81/2)(t-2.4)^2

When they meet, the two x(t)'s are equal, so
\[0.7+20t-(9.81/2)t^2 = 22.6-9.8(t-2.4)-(9.81/2)(t-2.4)^2\]
Solve for t to get the time at which the two balls meet.