A gull is flying horizontally 8.17 m above the ground at 6.35 m/s. The bird is carrying a clam in its beak and plans to crack the clamshell by dropping it on some rocks below. Ignoring air resistance, what is the horizontal distance to the rocks at the moment that the gull should let go of the clam?
how long does it take to drop the clam:
8.17=1/2 g t^2 solve for time, then
how far does it travel horiztohally?
distance=6.35*timeinair
To find the horizontal distance to the rocks at the moment the gull should let go of the clam, we need to determine how long it will take for the clam to reach the ground.
First, let's consider the vertical motion of the clam as it falls. We can use the equation of motion:
h = ut + (1/2)gt^2
Where:
h = height (8.17 m)
u = initial vertical velocity (0 m/s, as the clam is initially at rest)
g = acceleration due to gravity (-9.8 m/s^2, assuming downward as positive)
t = time
Rearranging the equation, we get:
t = √(2h / g)
Substituting the values, we have:
t = √(2 * 8.17 / 9.8) = √(16.34 / 19.6) ≈ 1.42 seconds
Since we are only concerned about the horizontal distance, we can now multiply the time by the horizontal velocity of the gull to find the distance:
distance = velocity * time
distance = 6.35 m/s * 1.42 s ≈ 9.02 meters
Therefore, the horizontal distance to the rocks at the moment the gull should let go of the clam is approximately 9.02 meters.