A clam is dropped by a seagull it takes 3 seconds to hit the ground. what is the seagulls height approximate height above the ground at the time the clam was dropped?
D=vit^2 + (1/2)at^2
D=(0m/s)(3s)^2 + (1/2)(9.8m/s^2)93s)^2
D=45m
Oh, I see we're doing some mathematical comedy here!
Well, to estimate the height, we can assume that the seagull wasn't feeling too energetic that day and just dropped the clam straight down. So, using the ol' Newtonian physics, we know that the distance an object falls can be calculated using the equation d = (1/2)gt^2, where d is the distance, g is the acceleration due to gravity (about 9.8 m/s^2 on Earth), and t is the time.
If we plug in the values, we have d = (1/2)(9.8 m/s^2)(3 s)^2, which simplifies to d ≈ 44.1 meters.
So, my approximate height estimate of the seagull above the ground when it dropped the clam is about 44.1 meters. Keep in mind, though, this is assuming the seagull didn't have any other ideas like going for a fly mid-drop!
To determine the approximate height above the ground at which the clam was dropped, we can use the equation for the distance an object falls due to gravity:
distance = (1/2) * acceleration due to gravity * time^2
Given that the time it takes for the clam to hit the ground is 3 seconds, we can substitute this value into the equation:
distance = (1/2) * 9.8 m/s^2 * (3 seconds)^2
Now we can calculate the distance:
distance = (1/2) * 9.8 m/s^2 * 9 seconds^2
distance = 44.1 meters
Therefore, the seagull was approximately 44.1 meters above the ground when it dropped the clam.