Physics, asked by shahdarsh601, 10 months ago

Pls answer the qn asap and also it's not there on any other website do don't go out searching and posting wrong and. Pls it's really imp

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Answered by harinarayanakancharl
0

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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