Solution needed : Kinematics problem

[dawoodvora]

Homework Statement

A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upward from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground.

Variables for Ball 1 going downward
v1_i = initial velocity of ball1
v1_f = final velocity of ball1

Variables for Ball 2 going downward
v2_i = initial velocity of ball2
v2_f = final velocity of ball2

t = time taken for both the balls to travel common meeting point
a = -9.8 m/s²

Homework Equations

4 general kinematic equations in one dimension having constant acceleration

The Attempt at a Solution

Logically, the final velocity of ball 1, when it reaches h/2, must also be the value of the initial velocity of ball2 going upward. Also, ball2 final velocity at h/2 must be zero. In this way, both the balls can meet each other at h/2

I am still unable to arrive at the figure of initial speed of ball2.

 

[kuruman]

The general way of doing this sort of problem is to
(a) First find the time it takes for the balls to pass each other by using the kinematic equation for the ball that is released from height h.
(b) Second use the same kinematic equation and the time you found in (a) to find the initial velocity for the ball that is projected up.
This method will allow you to solve the problem regardless of the fraction of h at which the two balls pass each other.

In this specific case, since you know that the speeds of the balls are the same when they pass each other, you can use the kinematic equation that does not involve time directly.

 

[Seannation]

EDIT: Disregard all of the below, see next post!

You're making innecessary assumptions here. The second ball's velocity needn't necessarily be zero at h/2. Since we're only dealing with one dimension here, there is only one possible initial speed of ball 2 that can allow it to hit ball 1 at h/2.

The two variables which will be equal for both balls when they hit is that they will both have traveled a distance of h/2. However one ball's displacement will be negative to the other as one ball is traveling down and one ball is traveling up.

You're looking to find the initial velocity of the second ball such that the distance traveled by the second ball against gravity equals the distance fallen by the second ball with gravity over the same time period.

 

[kuruman]

You're making innecessary assumptions here. The second ball's velocity needn't necessarily be zero at h/2.

If the balls meet halfway up, then the second ball's velocity is zero when they pass each other. This will not be the case if they met at some other fraction of h.

Proof
Let t_f = the time the balls meet. The average velocity for ball A is
\bar{v}_A=gt_f/2
The average velocity for ball B is
\bar{v}_B=(v_0+v_0-gt_f)/2=v_0-gt_f/2
where v₀ is what we are looking for.
Since the balls cover equal distances in time t_f, their average velocities must be the same.

gt_{f}/2=v_0-gt_f/2\ \rightarrow v_0=gt_f

Therefore the velocity of ball B at any time t is given by
v=gt_{f}-gt

At specific time t = t_f, ball B has zero velocity. Q.E.D.

 

[Seannation]

You make a very good point! I somehow got it in my head that the second ball would be traveling at high speed when it hit the first. But as you've proved, it can't have a non-zero velocity if it happens at h/2.