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Grade 12th passMechanics

A tower is 100m in height. A particle is dropped from the top of the tower and at the
same time another particle is projected upward from the foot of the tower. Both the
particles meet at height of 40m. Find the velocity with which the second particle is
projected upward

ram

8 Years agoGrade 12th pass

1 Answer

NIKHIL GHODKE

8 Years ago

this question can be spilt into two parts

calculate the time taken by the ball released from the height to meet with the second ball
using the the time calculated ,calculate the velocity of second ball

So how did i come to this conclusion

simple we are given that the balls meet at a point 40 m above ground ,so in this duration the released ball has covered (100-40)m right .SO BASICALLY U ARE GIVEN THAT THE TIME TAKEN BY THE RELEASED BALL TO COVER 60M FROM TOP OF BUILDING IS EQUAL TO TIME TAKEN BY SECOND BALL TO REACH 40 M FROM GROUND.

so now we calculate time taken by released ball to cover 60m

using

S=u*t+(a*t²)/2

here u=0

as initial it has no velocity

a= – gm/s2= – 10 m/s2

S= – 60m

(note in this sum i am taken scalars or vectors towards ground as -ve and away from ground as +ve)

(2S/a)^1/2=t

t=(120/10)^1/2sec

t=(12)^1/2

now using data calculate the second ball velocity

let the initial velocity be V

t=(12)^½sec

a= – g=

S=40m

S=V*t+(a*t²)/2

by substuting the values we get

V=50/(3)^1/2 m/sec

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