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Let us consider the speed of the ball thrown upwards as $U_1$ and the ball thrown downwards as $U_2$.

Then, $U_1 = 20m/s$ and $U_2 = 20m/s$

Consider X as the distance from the ground where the collision takes place then,

Distance from top of the tower up to the collision point = \[40{\text{ }}-{\text{ }}X\]

If it's suppose that collision took place after time T then

For upward motion, $X = 20T - \dfrac{1}{2} \times g \times {T^2}$ (Since the second equation of motion $S = ut + \dfrac{1}{2}a{t^2}$ )

For downward motion, $40 - X = 20T + \dfrac{1}{2} \times g \times {T^2}$

Adding both the equations, we get $40 = 40T$.

Which gives T = 1second

We can substitute the value of T in any of the two equations to get the answer.

Thus by substituting the value of T in the upward motion equation we get,

$X = 20(1) - \dfrac{1}{2} \times 9.8 \times {(1)^2}$ (Taking$(g = 9.8m/{s^2})$)

$X = 20 - 4.9 = 15.1\text{metre}$

So collision occurs at \[15.1\]metres from ground or \[40-15.1 = 24.9\] metres from top.

Assuming the reference frame of the ball A, From this reference frame, since Ball B and Ball A have the same acceleration ’g’ we need not to account for the acceleration.

For the balls to meet, the two must've covered a combined total of \[40\] metres, but from our Reference Frame, ball B must've covered these \[40\]metres with a speed in \[\left({20} + {20} \right)m/s\], or in $\dfrac{{40}}{{40}}$, or in 1 second. Hence by applying time as done in the solution we can find the collision distance.

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