Question

✨From the foot of tower 90m high a stone is thrown up so as to reach the top of Tower 2 second later another stone is dropped from the top of the tower when stone meet

Answer 1

ANSWERS :-


$solution$
Let the tour stone Mithi second after the projection of the first particle the sum of the distance moved by the particles is 90 m

h1 + h2 = 90 ...........(1)

Let u be wct of projection of the first particle as it recharge and only up to the top of the tower its velocity becomes zero,

so,

v² = u² - 2gh

0² = u² - 2 * g * 90

u² = 180g ,

u² = 180 * 9.8

u = √180 * 9.8 = 42 m/sec

Now, h1 = 42t - (1/2) * 9.8 * t²

h2 = (1/2) * 9.8 ( t -2 ) ² .........(ii)

substitute these values in equation (I) , we get

42t - 4.9t² + 4.9 ( t - 2 )²

= 42t - 19.6 t + 19.6 = 90

or,

t = 70.4/22.4 = 22/7 second

so,

h2 = 9.8/2 (22/7 -2)²

= 4.9 * 64/49

= 6.4 metre

h1 = 90 - 6.4 = 83.6 metre


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