A ball is allowed to fall from the top of the tower 200m high at the same instant another ball is thrown vertically upward from the bottom of the tower with a velocity of 40 meters per second?

Answer 1

Let us suppose that two balls meet after a time 't' at a height 'h' above the ground.

For the ball drop from the top:The ball will cover a distance of (200 - h) meters

Therefore, u = 0 ;

S = (200 - h) meters .

And, a =g= 9.8ms-2

So, S = ut + 1/2 at2

or, 200 - h = 0 x t +1/2 at2

or, 200 - h = 4.9t2 ...........................................(i)

For the ball thrown upwards :u = 40ms-1 ;

S = h;

And, a = g = -9.8ms-2 (-ve value of g since it is thrown upwards)

Again, S = ut + 1/2 at2

Therefore, h = 40 x t + 1/2(-9.8)t2

or, h = 40t - 4,9t2 ..............................................(ii)

Adding the equations (i) and (ii), we get

200 - h + h = 4.9t2 +40t - 4.9t2

Or, t = 5 seconds

Substituting for t = 5 seconds in (i), we have

200 - h = 4.9 x 52

Or, h = 200 - 4.9 x 52

= 200 - 122.5

= 77.5 meters

Therefore, the two balls will meet in 5 seconds and at the height of 77.5 meters........

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