Question

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Answer 1

Answer:

Given :-

Height of the tower = 30 m

Speed the ball thrown from the bottom is 30 m/s

To find :-

When will both the balls meet

How to find :-

Let's take that the ball thrown from the top of the tower (1 st particle) travels a distance (30 - x) and the ball dropped from the top (2 nd particle) travels x m.

Let both the balls meet at a time 't'.

For the first particle,

By applying $s = ut + \frac{1}{2} * a *t^{2}$

$(30 - x) = 0*t + \frac{1}{2} * g * t^{2}$

$(30 - x) = 5t^{2}$ ------ (1)

For the second particle,

By applying $s = ut + \frac{1}{2} * a *t^{2}$

$x = 30t - 5t^{2}$  ------ (2)  (Here we get -5 because the particle is moving opposite to the direction of g)

Adding both the equations we get,

100 = 30t

t = 10/3 s = 3.333 s

Now substituting t = 3.333 s in the first equation we get ,

$(30 - x) = 5*3.33^{2}$

30 - x ≈ 55.5 m

x = - 25.5 m (Negative sign indicates opposite to the motion of particle)

Hence both the particles meet at 25.5 m above the ground at 3.33 s.

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