A pelican flying along a horizontal path drops a fish from a height of 3.3 m. The fish travels 8.2 m horizontally before it hits the water below.
What was the pelican's initial speed? The acceleration of gravity is 9.81 m/s^.
Answer in units of m/s.
If the pelican was traveling at the same speed but was only 1.6 m above the water, how far would the fish travel horizontally before hitting the water below?
Answer in units of m.
Well, let's do some math while keeping it fishy! When the pelican drops the fish, it will travel in a projectile motion. We can use the equation of motion to solve this problem.
To find the pelican's initial speed, we can use the equation:
d = v0*t + (1/2)*a*t^2
Where d is the horizontal distance traveled (8.2 m), v0 is the initial speed we're looking for, t is the time it takes to reach the water, and a is the acceleration due to gravity (-9.81 m/s^2).
Since the fish is dropped from 3.3 m and we want to find the time it takes to reach the water, we can use the vertical motion equation:
d = v0*t + (1/2)*a*t^2
Substituting the values we have:
3.3 = 0*t + (1/2)*(-9.81)*(t^2)
Rearranging the equation, we get:
-4.905*t^2 + 3.3 = 0
Now it's just a little quadratic equation time! Solving for t, we find:
t ≈ 1.015 s
Now, let's plug this time back into the first equation to find the initial speed, v0:
8.2 = v0*(1.015) + (1/2)*(-9.81)*(1.015)^2
Solving for v0, we find:
v0 ≈ 10.04 m/s
So, the pelican's initial speed is approximately 10.04 m/s.
Now, onto the second part of the question! If the pelican is only 1.6 m above the water, we can use the same equation to find the horizontal distance traveled by the fish before it hits the water.
Substituting the new values:
1.6 = 0*t + (1/2)*(-9.81)*(t^2)
Solving for t, we find:
t ≈ 0.568 s
Now, using this time in the equation for horizontal distance:
d = v0*(0.568)
Plugging in the value for v0 from before, we get:
d ≈ 5.70 m
So, when the pelican is only 1.6 m above the water, the fish will travel approximately 5.70 m horizontally before it hits the water below.
Hope that answers your question with a splash of humor!
To find the initial speed of the pelican, we can use the equation of motion for the vertical displacement of the fish.
The equation is:
d = vit + 0.5at²
Where:
d = 3.3 m (vertical displacement)
vi = initial vertical velocity (which is equal to the initial speed of the pelican as the fish was dropped)
a = -9.81 m/s² (acceleration due to gravity)
t = unknown (time taken for the fish to hit the water)
Since the fish was dropped and not thrown, the initial vertical velocity of the fish is 0 m/s. Therefore, the equation becomes:
3.3 = 0 + 0.5(-9.81)t²
Simplifying the equation gives:
3.3 = -4.905t²
Rearranging the equation gives:
t² = 3.3 / -4.905
Calculating t gives:
t ≈ 0.804 seconds
The horizontal distance traveled by the fish is given by the equation:
x = vixt
Where:
x = 8.2 m (horizontal displacement)
vi = initial horizontal velocity (which is equal to the initial speed of the pelican as the fish was dropped)
t = 0.804 seconds (time taken for the fish to hit the water)
Rearranging the equation gives:
vi = x / t
Calculating vi gives:
vi = 8.2 m / 0.804 s
vi ≈ 10.199 m/s
Therefore, the pelican's initial speed was approximately 10.199 m/s.
For the second part of the question:
If the pelican was only 1.6 m above the water, we can use the same equation for horizontal distance to calculate how far the fish would travel.
Using the given values:
x = unknown (horizontal displacement)
vi = 10.199 m/s (initial horizontal velocity)
t = unknown (time taken for the fish to hit the water)
Rearranging the equation gives:
x = vit
Substituting the values gives:
x = 10.199 m/s * t
From the first part of the question, we know that the time taken for the fish to hit the water is approximately 0.804 seconds. Substituting this value into the equation gives:
x = 10.199 m/s * 0.804 s
Calculating x gives:
x ≈ 8.214 m
Therefore, if the pelican was only 1.6 m above the water, the fish would travel approximately 8.214 m horizontally before hitting the water below.
To find the initial speed of the pelican, we can use the equations of motion. Since the pelican dropped the fish from a height of 3.3 m and the fish traveled 8.2 m horizontally, we can assume that the time it took for the fish to hit the water is the same as the time it took for the fish to travel horizontally.
Using the equation for vertical displacement, we have:
h = (1/2)gt^2
Where h is the initial height, g is the acceleration due to gravity, and t is the time.
For the fish, h = 3.3 m. Plugging in values, we can solve for t:
3.3 = (1/2)(9.81)t^2
6.6 = (9.81)t^2
t^2 = 6.6/9.81
t ≈ 0.879 sec (approximately)
Now, using the equation for horizontal displacement, we have:
d = vxt
Where d is the horizontal distance, v is the initial velocity, and t is the time.
For the fish, d = 8.2 m and t ≈ 0.879 sec. Plugging in values, we can solve for v:
8.2 = v(0.879)
v ≈ 9.33 m/s (approximately)
Therefore, the pelican's initial speed was approximately 9.33 m/s.
Now, let's calculate how far the fish would travel if the pelican was only 1.6 m above the water. Similarly, we'll find the time it takes for the fish to hit the water from this lower height.
Using the same equation for vertical displacement, h = 1.6 m. Plugging in values, we can solve for t:
1.6 = (1/2)(9.81)t^2
3.2 = (9.81)t^2
t^2 = 3.2/9.81
t ≈ 0.567 sec (approximately)
Now, using the same equation for horizontal displacement, we can solve for d:
d = vxt
For the fish, t ≈ 0.567 sec. Plugging in the value of t and the initial velocity v (which we found to be approximately 9.33 m/s), we can solve for d:
d = (9.33)(0.567)
d ≈ 5.285 m (approximately)
Therefore, if the pelican was only 1.6 m above the water, the fish would travel approximately 5.285 m horizontally before hitting the water.
1. Dx = Xo*Tf = 8.2 m.
h = 0.5g*Tf^2 = 3.3 m.
4.9*Tf^2 = 3.3.
Tf^2 = 0.673.
Tf = 0.821 s. = Fall time.
Xo*0.821 = 8.2 m.
Xo = 10.0 m/s.
2. Dx = Xo*Tf.
h = 0.5g*Tf^2 = 1.6 m.
4.9*Tf^2 = 1.6.
Tf^2 = 0.327.
Tf = 0.571 s.
Dx = 10m/s * 0.571s. = 5.71 m.