boy throws a penny into an empty well with an initial velocity of –2 m/s. He hears it hit the bottom 8 seconds later. About how far did the penny fall?
t'=time to fall
t"=time for sound to return after bottom
t'+t"=8
depth=2*t'+1/2 g t'^2
depth=velocitysound*t"
set them equal (same depth)
2*t'+1/2 g t'^2=velocitysound*t"
2*t'+1/2 g t'^2=velocitysound*(8-t')
so solve for t', and then you can get depth.
To find the distance the penny fell, we can use the equation of motion for a falling object:
d = vit + (1/2)gt^2
where:
- d is the distance
- vi is the initial velocity
- t is the time
- g is the acceleration due to gravity
In this case, the initial velocity of the penny is -2 m/s (negative because it is thrown downwards), and the time is 8 seconds.
Acceleration due to gravity, g, is approximately 9.8 m/s^2.
Let's substitute the values into the equation and calculate the distance:
d = (-2 m/s)(8 s) + (1/2)(9.8 m/s^2)(8 s)^2
Simplifying the equation:
d = -16 m + (1/2) * 9.8 m/s^2 * 64 s^2
d = -16 m + 313.6 m
d = 297.6 m
Therefore, the penny fell approximately 297.6 meters.