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Standard XII

Physics

Distance

A ball is dro...

Question

A ball is dropped from a platform at the=0 starting from rest after 6 sec another ball is thrown downwards from the same platform with speed v the two balls meet at time 18 sec what is value of v

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Solution

diatance travelled by 1st ball in 18 sec,

s=vt+(1/2)at^2

s=0+(1/2)×10×18×18

s=90×18s=1620m

time taken by second ball=18-6=12sec

so,distance travelled by second ball

s=vt+(1/2)gt×t

so,1620=12×v+(1/2)×10×12×12

1620=12×v+1440

1620=12×v+720

12×v=900

V=75m/s

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A ball is dropped from a platform at the=0 starting from rest after 6 sec another ball is thrown downwards from the same platform with speed v the two balls meet at time 18 sec what is value of v

diatance travelled by 1st ball in 18 sec, s=vt+(1/2)at^2 s=0+(1/2)×10×18×18 s=90×18s=1620m time taken by second ball=18-6=12sec so,distance travelled by second ball s=vt+(1/2)gt×t so,1620=12×v+(1/2)×10×12×12 1620=12×v+1440 1620=12×v+720 12×v=900 V=75m/s

diatance travelled by 1st ball in 18 sec,

s=vt+(1/2)at^2

s=0+(1/2)×10×18×18

s=90×18s=1620m

time taken by second ball=18-6=12sec

so,distance travelled by second ball

s=vt+(1/2)gt×t

so,1620=12×v+(1/2)×10×12×12

1620=12×v+1440

1620=12×v+720

12×v=900

V=75m/s