a stone is dropped from the top of a tower 500m high into a pond of the tower. when is the splash herd at the top? given, g= 10m/s2 and speed of sound=340m/s.
Answers
Answer:
Height of the tower, s = 500 m
Velocity of sound, v = 340 m sˆ’1
Acceleration due to gravity, g = 10 m sˆ’2
Initial velocity of the stone, u = 0 (since the stone is initially at rest)
Time taken by the stone to fall to the base of the tower, t1
According to the second equation of motion:
S = ut1 + 1/2gt12
500 = 0 x t1 + 1/2 x 10 x t12
t12 = 100
t1 = 10s
Now, time taken by the sound to reach the top from the base of the tower, t2= 500/340 = 1.47 s
Therefore, the splash is heard at the top after time, t
Where, t= t1 + t2 = 10 + 1.47 = 11.47 s.
CORRECT QUESTION :-
- A stone is dropped from the top of the tower 500m high into the pond of water at the base of tower. When is the splash heard at the top of the tower. Given, g=10m/s2 and speed of sound is 340m/s.
SOLUTION :-
- Distance (s) = 500 m
- Acceleration due to gravity (g) = 10 m/s²
- Speed of sound (v) = 340 m/s
- Initial velocity (u) = 0 m/s
★ By using second equation of motion we get :
- s = ut + ½ gt²
★ Plug in given values in above equation of motion :
- → 500 = 0 × t + ½ × 10 × (t)²
- → 500 = ½ × 10 × t²
- → 500 = 5 × t²
- → t² = 500 ÷ 5
- → t² = 100
- → t = √100
- → t = 10 seconds
★ Now, let's calculate time taken by the sound to reach the top from the base of the :
→ t' = Distance ÷ speed of sound
- → t' = 500 ÷ 340
- → t' = 50 ÷ 34
- → t' = 1.47 seconds
★ The splash heard at the top of the tower :
- → Total time = t + t'
- → Total time = 10 + 1.47
- → Total time = 11.47 seconds