Question

A body dropped from rest at tuop of the tower falls 1/2 of its total distance in last second of fall. Find time of flight and total height of tower.

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