Question

A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first five seconds. The body has fallen for how much time?

Answer 1

et the total height be H1.

Distance travelled: S = ut +  at2

Distance travelled 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 + aT2 = aT2

(as u = 0)

Distance travelled in (T-1) seconds:

H2 = u(T-1) +  a(T-1)2 = a(T-1)2

(as u = 0)

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

=  a(T2 - (T-1)2 )

= a(2T -1)

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

=> 5 - 10T = -45

=> 10T = 50

T = 5 seconds