An eagle flying along a horizontal path drops a small snake from a height of 6.5 m while travelling 5.0 m/s. How far does the snake travel horizontally before it hits the ground below?
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An eagle flying along a horizontal path drops a fish from a height of 5.4 meter. The fish travels 8.0 meter horizontally before it hits the water below. What is the eagle initial speed
G: Why did the eagle drop the small snake? Was it trying to slither out of paying the eagle's taxi fare?
R: Alright, let's calculate how far the snake travels horizontally before it hits the ground.
E: We can use the formula for horizontal distance traveled: d = v * t, where v is the horizontal velocity of the snake and t is the time it takes for the snake to hit the ground.
S: To find the time, we need to figure out how long it takes for the snake to reach the ground. We can use the equation h = 0.5 * g * t^2, where h is the initial height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
A: Plugging in the values, we have 6.5 = 0.5 * 9.8 * t^2. Solving for t, we get t = √(6.5 / (0.5 * 9.8)), which is approximately 0.9 seconds.
Now that we have the time, we can calculate the horizontal distance using d = v * t. With a velocity of 5.0 m/s and a time of 0.9 seconds, the snake would travel approximately 4.5 meters horizontally before slithering to the ground.
G: The given values in the problem are the height from which the snake is dropped (6.5 m) and the horizontal velocity of the eagle (5.0 m/s). We need to find out the horizontal distance traveled by the snake before it hits the ground.
R: To solve this problem, we need to split it into two separate components - the horizontal component and the vertical component. The horizontal component is given by the velocity of the eagle, which is 5.0 m/s. The vertical component is given by the height from which the snake is dropped, which is 6.5 m.
E: We can use the equation of motion to calculate the time it takes for the snake to fall to the ground. The equation we can use is:
h = (1/2)gt^2
Where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. Plugging in the values, we have:
6.5 = (1/2) * 9.8 * t^2
Solving for t, we get:
t = √(13/9.8)
Now that we have the time, we can use the horizontal velocity to calculate the horizontal distance traveled by the snake:
d = v * t
where d is the distance, v is the horizontal velocity (5.0 m/s), and t is the time. Plugging in the values, we have:
d = 5.0 * √(13/9.8)
S: The snake travels approximately 6.37 meters horizontally before it hits the ground below.
A: The snake travels approximately 6.37 meters horizontally before it hits the ground below.
distance=5.0*time
where time=sqrt(2*5/9.8) second