Question
Mon May 09, 2011 By:

Tue May 10, 2011
When we throw a body upward, the acceleration is against the gravity so it is taken as minus therefore the equation which we know comes out to be

, and when ut is less than , then distance comes out to be negative sign.

For the problem:

Soln: The displacement of the body in first 2 sec =200cm,

Let the initial velocity = u(say) ,

Acceleration = a(say), time  =2sec

Substitute the value of  in above equation, we get

= 2u +2a ; 2(u+a) =200

Therefore u+a = 100 Â– - Â– - Â– - Â– - Â– - Â– - - Â– - Â– - - Â– (1)

Given that the body travels 220cm in next 4sec.That is from the start it displaces 200+220 = 420 cm in 6sec.

Displacement =420 cm,

time =6 sec,

substitute theses values in the equation ,

6(u+3a) = 420 ; u+3a = 70 Â– - Â– - Â– - Â– - Â– - Â– - - Â– - Â– - - Â– (2)

Solving equations (1) and (2) or Eq (2) Â– Eq(1)

we get 2a= -30 ; a=-15 cm/,

Substitute value of a in Eq(1) we get u-15 = 100,

u = 115 cm/sec.

The velocity at the end of second v=u+a

we get v= 115+(-15)(6) =115-90 =25cm/sec.

Therefore final velocity v=25cm/sec.
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