Question

two bodies are held seperated by 9.8m vertically one above the other.they are released simuntaneously to fall freely under gravity.after 2 second distance between them is ?

Answer 1

Answer:

FOR OBJECT 1

Initial velocity=x

acceleration=10(easy calculations)

Time=2

S=1/2at^2+ut

S=1/2*10*4+2x

S1=20+2x

FOR OBJECT 2

initial velocity=x+9.8

Acceleration =10

Time=2

S=1/2at^2+ut

S=1/2*10*4+2(x+9.8)

S2=20+2x+19.6

DISTANCE BETWEEN BOTH THE OBJECTS

S2-S1

20+2x+19.6-(20+2x)

=20+2x+19.6-20-2x

=19.6

SO THE DISTANCE BETWEEN THE 2 OBJECTS AFTER 2 SECONDS WILL BE 19.6 m

Answer 2

Explanation:

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