Question : A ball is dropped from a high tower such that distance covered by it in last second of its motion is same as the distance covered by it during first three seconds. Find the time taken by ball to reach ground and height of tower. (Take g=10 ms2)
Doubt by Japjee and Krish
Solution :
Let the total time taken by the body to reach the foot of the tower = t
and the height of the tower = h
As per the question
Distance covered by the body in the first three seconds = Distance covered by the body in the last second.
s3=st
(0)(3)+½(+g)(3)2 = (0)+½(g)[2(t)-1]
(0)(3)+½(+g)(3)2 = (0)+½(g)[2(t)-1]
[Using the relation, s=ut+½at2 and snth = u+½g(2n-1), putting n=t]
½(+g)(3)2 = ½(g)[2t-1]
(3)2 = [2t-1]
9=2t-1
9+1=2t
9+1=2t
10=2t
10/2=t
t=5 sec
Now, total distance covered by the body in 5 sec will give us the total height of the tower
s=ut+½gt2
s=ut+½gt2
s=0(4)+½(+10)(5)2
s=0+5(25)
s=125 m
Hence, time taken by the ball to reach the ground is 5 sec and height of the tower is 125 m