A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of the tower is
(g=10 m/s2)
Let the body fall through the height of the tower in t seconds.
From equation of motion we have
sn=u+a2(2n−1)
So, we have total distance travelled in last 2 seconds of fall as
40=st+s(t−1)
⇒ 40=[0+g2(2t−1)]+[0+g2(2t−3)]
⇒ 40=g2(4t−4)⇒ t=3 sec
Thus, distance travelled in 3 sec will be the height of the tower which is given by
h=ut+12gt2=0+12×10×32
⇒ h=45 m