signs for free falling bodies
Asked by yashpinto
| 9th May, 2011,
05:24: PM
Expert Answer:
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 - - - - - - - - - - (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 - - - - - - - - - - (2)
Solving equations (1) and (2) or Eq (2) 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.



Let the initial velocity = u(say) ,
Acceleration = a(say), time


Substitute the value of



Therefore u+a = 100 - - - - - - - - - - (1)
Given that the body travels 220cm in next 4sec.That is from the start it displaces 200+220 = 420 cm in 6sec.
Displacement

time

substitute theses values in the equation


6(u+3a) = 420 ; u+3a = 70 - - - - - - - - - - (2)
Solving equations (1) and (2) or Eq (2) 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.
Answered by
| 10th May, 2011,
07:24: PM
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