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Grade: 11
A ball is dropped from the top of a tower of height 100m and at the same time another ball is projected vertically upwards from ground with a velocity 25 m/s. Then distance from the top of the tower ,at which the two balls meet is
9 months ago

Answers : (2)

Vikas TU
11143 Points
Dear Student 
Lets take distance covered by the stone thrown upwards to meet the other as x. Then the other stone covers a distance of (100-x)m.
Let the ball dropped down be b1. So the ball thrown up is b2.
b1 => d = 100-x
          g = 9.8 m/s sq.
          u = 0
We know, s= ut + 1/2at sq.
Putting values,
100-x = 4.9t sq.                  ....(1)
b2 => d = x
          g = -9.8 m/s sq.
          u = 25 m/s 
We know, s= ut + 1/2at sq.
Putting values,
x = 25t -4.9t sq.                  ....(2)
Adding (1) and (2),
100-x +x = 4.9t sq. + 25t - 4.9t sq.
t = 4 secs.
Puuting t = 4 in (1),
100-x = 4.9t sq.
100-x =19.6
x = 80.4 m
So, they meet at  80.4 m from ground after 4 seconds.
9 months ago
3008 Points

Each ball is moving in a straight line under uniform acceleration and is thus subject to the equation:

S = ut + 1/2 a t^2

… where S is distance travelled, a is the acceleration towards the ground and t is the journey duration.

We know the sum of their journeys is 100m and their journey durations are equal.


100 = (25t -1/2 g t^2) + (0t + 1/2 g t^2)

100 = 25t

So t = 4 seconds

Note: This is is the same time that it would have taken had there been no gravity and the thrown ball covered the entire distance itself. That is intuitive as whatever retarding gravity applies to the thrown ball is equal to the speeding up it applies to the dropped ball.

Note 2 (thanks to comment by Victor Mazmanian): We can assess when the thrown ball reaches its maximum height by solving (v = u+ at) for t when v is 0.

0 = 25 - gt

So t is 25/g … roughly 2.5 seconds

Hence at 4 seconds, when the balls meet the thrown ball is already on its way down after already reached its max height and started descending.

Under the effects of gravity the thrown ball reaches a max height of…

25.25/g - 1/2.g.(25/g)^2 = roughly 31.9 m

but has descending back down to a height of …

25.4 - 1/2.g.4^2 = 100–8g … roughly 21.5m, when the balls meet

And the dropped ball this covers the remaining 1/2.g.4^2 = 78.5m

9 months ago
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