Question

Drops of water fall at regular intervals from the roof of a building of height H = 16 m , the first drop strikes the ground at the same moment when the fifth drop detaches itself from the roof. The distances between the different drops in air as the first drop reaches the ground are :

Answer 1

"The distances travelled by the second, third, and fourth drops are 9m, 4m, and 1m respectively.

Given:

Height of the building, s = 16 m

Let $\mathrm{s}_{2}, \mathrm{s}_{3}, and \mathrm{s}_{4}$ indicate the distances of second, third, and fourth drop from the roof respectively.

Let t indicate the time taken by the drop to reach the ground at regular interval.

Since, the drops fall at regular intervals, one can say that the first drop takes 4t seconds to reach the ground.

$\mathrm{s}_{2}=?$

$\mathrm{s}_{3}=?$

$\mathrm{s}_{4}=?$

Solution:

The distance travelled by a body undergoing free fall motion is

$s=\frac{g t^{2}}{2}$

On substituting, we get

For the first drop,

$16=\frac{9.8 \times(4 t)^{2}}{2}$

$t=\sqrt{\frac{1}{4.9}}$

For the second drop,

$s_{2}=9.8 \times \frac{\left[3 \times \sqrt{\frac{1}{4.9}}\right]^{2}}{2}$

$s_{2}=\frac{9.8 \times 9}{2 \times 4.9}=9 m$

For the third drop,

$s_{3}=9.8 \times \frac{\left[2 \times \sqrt{\frac{1}{4.9}}\right]^{2}}{2}$

$s_{3}=\frac{9.8 \times 4}{2 \times 4.9}=4 m$

For the fourth drop,

$s_{4}=9.8 \times \frac{\left[1 \times \sqrt{\frac{1}{4.9}}\right]^{2}}{2}$

$s_{4}=\frac{9.8 \times 1}{2 \times 4.9}=1 m$"

Answer 2

Explanation:

given the height of building =16m.

the given condition is when the first drop strikes the ground at the same time the 5 drop leaves the roof

let the distance between each drop is "s"

1drop =S1

2drop =S2

3drop =S3

4drop=S4

5drop=S5

we need to before that for a freely falling bodies the initial velocity=0.

and the distance between them is "t".

for 5drop is t

4 drop is 2t

3 drop is 3t

2 drop is 4t seconds.

S=U+1/2gt^2_______________________ 1

u=0

so s=1/2gt^2

S4=1/2gt^2

s3= 1/2g(3t^2)

2s2=1/2g(4t^)

s1=1/2g(4t^2)

by the given condition

s5 = 0

because it is on the roof.

we need subtract s4_s5.because the distance traveled by s4 is more than s5.

s1=16

1/2gt^2=16

1/2gt^2=16/16

1/2gt^2 =1_____________________ 2

s4_s5

1/2gt^2_0

1/2gt^2 =1 (from equation 2)

s3 _s4

1/2g(2t)^2_1/2gt^2

1/2 gt^2 gets cancelled

we get 4_1 =3m

s2_s3

1/2g(3t)^2_1/2g(2t)^2

9_4=5m

s1_s2

1/2g(4t)^2_1/2g(3t)^2

16_9

=7m