Question

A stone is dropped from the roof of a high building. A second stone is dropped 1.04 s later.

How far apart are the stones when the second one has reached a speed of 14.7 m/s?
Express your answer to three significant figures and include the appropriate units.

Answer 1

Step-by-step explanation:

Let the initial speed of 2nd stone is u2 and final speed of it be v2.

Using the equation

v= u + at for the 2nd stone

We get

14.7=u2 + 10t (a=g, acceleration due to gravity)

14.7=0 +10t

Or, t= 1.47s

This is the time taken by 2nd stone to reach 14.7m/s

Now,

Total time for 1st stone=1.04 + 1.47= 2.51s

Using the equation

`s = ut + (1)/(2) at {}^(2)`

for distance travelled by 1st stone

s1=u2t + 1/2at²

u2=0 (initially it was also at rest)

`s1 = (1)/(2) * 10 * 2.51 {}^(2)`

therefore S = 31. 5005m

Distance travelled by 2nd stone

`s2 = u2 + (1)/(2) at {}^(2)`

`s2 = (1)/(2) * 10 * 1.47 {}^(2)`

therefore S2 = 10.8m

Therefore,

Distance between the stones is= s1 - s2 = 31.5005–10.8 = 20.7005

Ans:- 20.7m