a stone is dropped from top of a tower 500 hai into a pond of water at the base of the tower . when is the splash held at the top given g is equal to 10 metre per second to the power minus 2 and speed of sound is 340 metre per second to the power minus one
Answers
Explanation:
a stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m s-2 and speed of sound = 340 m s-1. Where, t= t1 + t2 = 10 + 1.47 = 11.47 s.
Answer
Given ,
- Distance covered by the stone , (s)= 500m
- initial speed ,(u) = 0
- Acceleration ,(g) = 10 m s^-1
To find ,
- The time taken (t) = ?
- and the speed of sound in the top of tower = ?
Now ,
- by using the formula s = ut + 1/2 gt^2 we can find out the value of time taken , As here we are having the distance covered , initial speed and acceleration . So, we use this formula .
So ,
- putting the values in equation we get ;
=> 500 = 0(t) +1/2(10)(t^2)
=> 500 = 0 + 1/2(10)(t^2)
=> 500 = 1/2(10)(t^2)
=> 500 = 5 (t^2)
=> t^2 = 500/5
=> t^2 = 100
=> t = √100
=> t = 10s
the time taken is 10 second .
Thus , the stone take 10 second to fall into pound of water and produce the sound of splash. this sound of class has to travel a distance of 500 to be held at the top of Tower.
Now ,
Speed of sound = distance travelled by the sound / time taken by the sound
So ,
=> 30 = 500/ time taken by the sound
And ,
=> time taken by sound = 500/340 s = 1.47 s
Time after which Splash is Heard In top of the tower = 10s + 1.47s
time after which Splash is heard in the top of the tower = 11.47s .