Question

a body Falls freely from rest it cover as much distance in last second of its motion as cover in the first three second the body fallen for time of?
ans:5sec​

Answer 1

Let the total height be H1.

Distance traveled: S = ut +  1/2 at²

Distance traveled in the first three seconds: (put u=0, a = -10, t = 3 above):

S = - 45m

Let the total time be T. Therefore, distance travelled in complete T seconds:

H1 = uT + 1/2 aT²= 1/2 aT²

(as u = 0)

Distance travelled in (T-1) seconds:

H2 = u(T-1) +  1/2 a(T-1)²= 1/2 a(T-1)²

(as u = 0)

Distance travelled in last second of its motion: H1 - H2

=  1/2 a(T2 - (T-1)2 )

=1/2 a(2T -1)

Putting a = -10 and equating it to (-45) from above:

=> 5 - 10T = -45

=> 10T = 50

T = 5 seconds

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Answer 2

Answer:

Answer is 5 sec.

Explanation:

Distance covered in first 3 sec.

S = 1/2gt^2 = 1/2x9.8x3x3

distance covered in last second

S =9/2(2t-1)

1/2x9.8x9 =9/2(2t-1)

t = 5.4

t is nearly about 5

therefore t = 5sec.