Question

Drops of water fall at regular intervals from the roof of a building of height 16 m. The first drop strike the ground at the same moment as the fifth drop leaves the roof. Find the distance between the successive drops.

Answer 1

Hii dear,

# Answer-
Time taken by 1st drop to to reach ground = t
s = ut + 1/2 at^2
We have s=16m, u=0m/s, a=g,
16 = 0×t + 0.5×10×t^2
t^2 = 3.2
t = 1.8 s

a) For first ball,
s = 1/2×10×(t)^2
s = 16m

b) For second ball,
s = 1/2×10×(3t/4)^2
s = 9m

b) For third ball,
s = 1/2×10×(t/2)^2
s = 4m

b) For fourth ball,
s = 1/2×10×(t/4)^2
s = 1m

b) For fifth ball,
s = 1/2×10×(0)^2
s = 0m

(Note-All distances are from top.)

Hope this helps you...

Answer 2

Given :

Height=h=16m

Time taken by the first drop to touch the ground =t=√2h/g

=√2x16/9.8

=√3.26 sec

=1.8 s

Time interval between two successive drops is

= t/ (n-1)

Where n is number of drops

=1.8/5-1=1.8/4=0.45 sec

For second drop= h2=( ½) gt2²

=1/2 x 9.8 x 1.35 x 1.35

=8.93 m

D12=16- 8.93= 7.06=7m

For 3rd drop : h3=( ½) x (9.8 x 0.90x0.90)

=3.97

D23=8.93-3.97=4.961=5m

For 4th drop: h4= ½ x 9.8 x 0.45 x0.45

=0.9922

D34=3.97-0.9922

=2.9778=3m

Similarly for d45=0.9922-0=0.9922=1m