If the pelican was traveling at the same speed
but was only 2.3 m above the water, how
far would the fish travel horizontally before
hitting the water below?
Answer in units of m
To find the horizontal distance the fish would travel before hitting the water, we need to use the equations of motion and apply the concept of projectile motion.
First, let's assume that the initial vertical velocity of the fish is zero (since it is dropped into the water) and the only force acting on it is gravity. The horizontal velocity remains unchanged throughout the motion.
The equation we can use to solve this problem is the kinematic equation for vertical motion:
Δy = v₀y * t + (1/2) * a * t²
Where:
Δy is the displacement in the vertical direction (negative since the fish is moving downwards),
v₀y is the initial vertical velocity (zero in this case),
a is the acceleration due to gravity (approximately -9.8 m/s²),
t is the time of flight.
We can solve this equation for t:
Δy = (1/2) * a * t²
2.3 m = (1/2) * (-9.8 m/s²) * t²
t² = (2.3 m) / (1/2 * -9.8 m/s²)
t² = -0.469 m /s²
t = √(-0.469 m /s²)
Since time cannot be negative, we know that the fish will hit the water after a certain positive time. Now, we can use the horizontal velocity of the fish to calculate the horizontal distance traveled during this positive time.
The horizontal distance traveled (Δx) can be calculated using the equation:
Δx = v₀x * t
Where:
Δx is the horizontal distance traveled,
v₀x is the initial horizontal velocity (same as the pelican's speed),
t is the time of flight (as calculated earlier).
Since the initial horizontal velocity is equal to the pelican's speed, Δx can be calculated using:
Δx = speed * t
Now, you can substitute the positive value of t calculated earlier and the given speed of the pelican into this equation to find the horizontal distance traveled by the fish before hitting the water.
To determine the horizontal distance the fish would travel before hitting the water, we need to calculate the time it takes for the fish to reach the water.
The vertical distance traveled by the fish can be found using the equation of motion:
Δy = V₀yt + (1/2)gt²
Where:
Δy = vertical distance traveled (2.3m)
V₀y = initial vertical velocity (0m/s, since the fish was dropped)
g = acceleration due to gravity (-9.8m/s²)
t = time taken
Rearranging the equation, we get:
2.3 = 0t + (1/2)(-9.8)t²
2.3 = -4.9t²
t² = 2.3 / -4.9
t² ≈ -0.4694
(taking the square root of both sides)
t ≈ ±√(-0.4694)
The square root of a negative number is not a real number, so we discard the negative solution. Therefore, the time taken by the fish to hit the water is approximately zero.
Since the time is very close to zero, we can assume that the horizontal motion of the fish is unaffected. Thus, the horizontal distance it would travel before hitting the water would be the same as when dropped from 4.9m above the water.
Therefore, the fish would travel approximately 4.9 meters horizontally before hitting the water.