In order to open the clam it catches, a seagull will drop the clam repeatedly onto a hard surface from high in the air until the shell
cracks.
If a seagull flies to a height of 28.7 m, how
long will the clam take to fall? The accelera-
tion of gravity is 9.8 m/s2 .
Answer in units of sec
To find the time it takes for the clam to fall, we can use the formula for the time of free fall:
t = sqrt(2h/g)
Where:
t = time (in seconds)
h = height (in meters)
g = acceleration due to gravity (9.8 m/s^2)
Given that the seagull flies to a height of 28.7 m and g = 9.8 m/s^2, we can substitute these values into the equation:
t = sqrt(2 * 28.7 / 9.8)
t = sqrt(57.4 / 9.8)
t = sqrt(5.857)
Calculating the square root, we find:
t ≈ 2.42 seconds
Therefore, the clam will take approximately 2.42 seconds to fall.
To calculate the time it takes for the seagull to drop the clam from a height of 28.7 m, we can use the equation of motion:
h = (1/2) g t^2
Where:
- h is the height (28.7 m in this case),
- g is the acceleration due to gravity (9.8 m/s^2), and
- t is the time taken to fall.
We need to rearrange the equation to solve for t. First, let's eliminate the fraction by multiplying both sides by 2:
2h = g t^2
Next, we can rearrange the equation to isolate t^2:
t^2 = (2h) / g
Now we can take the square root of both sides to solve for t:
t = sqrt((2h) / g)
Plugging in the given values:
t = sqrt((2 * 28.7) / 9.8)
Calculating this expression:
t ≈ 2.85 seconds
Therefore, it will take approximately 2.85 seconds for the clam to fall from a height of 28.7 m.