Suppose a chinook salmon needs to jump a waterfall that is 1.32 m high.
(a) If the fish starts from a distance 1.24 m from the base of the ledge over which the waterfall flows, find the x- and y-components of the initial velocity the salmon would need to just reach the ledge at the top of its trajectory.
v0x = m/s
v0y = m/s
Kinda depends on the angle.
In x: 1.24=v cos(theta)t
In y: 1.32 = v sin(theta)t - 1/2(9.8)t^2
Stuck with three unknowns and only two equations.
To find the x- and y-components of the initial velocity, we can use the equations of motion.
Let's assume the acceleration due to gravity is g = 9.8 m/s².
(a) Finding the x-component of the initial velocity (v0x):
The horizontal motion is unaffected by gravity, so the x-component of the initial velocity remains constant throughout the motion.
The distance (d) from the base of the ledge to the top can be calculated by using the Pythagorean theorem:
d = √(1.24² + 1.32²)
d = √(1.5376 + 1.7424)
d = √3.28
d ≈ 1.81 m
The time taken to travel the horizontal distance (d) can be found using the equation:
d = v0x * t
Solving for t:
t = d / v0x
Now, we know that the time taken to reach the maximum height is half the total time of the motion, so the total time (t_total) is:
t_total = 2 * t
The maximum height (h_max) can be calculated using the equation of motion:
h_max = v0y * t - (1/2) * g * t²
We want the salmon to just reach the ledge at the top of its trajectory, which means the height at the top should be equal to the height of the ledge:
h_max = 1.32 m
Now, we can substitute the known values and solve for v0x and v0y:
1.32 = v0y * t - (1/2) * g * t²
Substituting t = d / v0x and t_total = 2 * t:
1.32 = v0y * (d / v0x) - (1/2) * g * (d² / v0x²)
Rearranging the equation:
1.32 = (v0y / v0x) * d - (1/2) * g * (d² / v0x²)
Since we are given that the fish just reaches the ledge, the maximum height should be equal to the height of the ledge:
1.32 = (v0y / v0x) * 1.81 - (1/2) * g * (1.81² / v0x²)
Now, we have two unknowns (v0x and v0y) in one equation. To solve for them, we need a second equation. The second equation can be formed by ensuring the vertical distance (h_total) traveled by the salmon is zero at the end of motion:
0 = v0y * t_total - (1/2) * g * t_total²
Substituting t_total = 2 * t:
0 = v0y * (2 * t) - (1/2) * g * (2 * t)²
0 = 2 * v0y * t - g * 4 * t²
Dividing both sides by 2t:
0 = v0y - 2 * g * t
Rearranging the equation:
v0y = 2 * g * t
Substituting t = d / v0x:
v0y = 2 * g * (d / v0x)
Now, we can substitute this value of v0y into the first equation:
1.32 = (2 * g * (d / v0x) / v0x) * d - (1/2) * g * (d² / v0x²)
Simplifying the equation:
1.32 = (2 * g * d² - (1/2) * g * d²) / v0x²
Cross-multiplying:
v0x² = (2 * g * d²) / (1.32 + 0.5 * g * d²)
Taking square roots:
v0x = √[(2 * g * d²) / (1.32 + 0.5 * g * d²)]
Substituting the known values:
v0x = √[(2 * 9.8 * 1.24²) / (1.32 + 0.5 * 9.8 * 1.24²)]
Calculating v0x:
v0x ≈ √(29.76 / 7.7656)
v0x ≈ √3.833
v0x ≈ 1.958 m/s
Now, substituting this value of v0x back into the equation for v0y:
v0y = 2 * g * (d / v0x)
v0y = 2 * 9.8 * (1.81 / 1.958)
Calculating v0y:
v0y ≈ 17.96 / 1.958
v0y ≈ 9.186 m/s
Therefore, the x- and y-components of the initial velocity the salmon would need to just reach the ledge at the top of its trajectory are:
v0x ≈ 1.958 m/s
v0y ≈ 9.186 m/s
To find the x- and y-components of the initial velocity (v0x and v0y) that the salmon would need to just reach the ledge at the top of its trajectory, we can use the principles of projectile motion.
Projectile motion can be analyzed by splitting the motion into horizontal (x) and vertical (y) components.
Given:
Height of the waterfall (h) = 1.32 m
Distance from the base of the ledge (x) = 1.24 m
We can start by finding the time it takes for the salmon to reach its highest point (t) using the equation:
h = (1/2) * g * t^2
Where g is the acceleration due to gravity, approximately 9.8 m/s^2 on Earth.
Rearranging the equation, we can solve for t:
t^2 = (2 * h) / g
t = sqrt((2 * h) / g)
Substituting the given value for h, we have:
t = sqrt((2 * 1.32) / 9.8) = sqrt(0.269) ≈ 0.519 s
Now, to find the x-component of the initial velocity (v0x), we can use the equation:
x = v0x * t
Rearranging the equation, we have:
v0x = x / t
Substituting the given value for x and the calculated value for t, we have:
v0x = 1.24 m / 0.519 s ≈ 2.390 m/s
Therefore, the x-component of the initial velocity required for the salmon to just reach the ledge is approximately 2.390 m/s.
For the y-component of the initial velocity (v0y), we can use the equation of motion:
y = v0y * t + (1/2) * g * t^2
Since the salmon's displacement in the y-direction is zero when it reaches the highest point, the equation becomes:
0 = v0y * t - (1/2) * g * t^2
Rearranging the equation, we have:
v0y * t = (1/2) * g * t^2
v0y = (1/2) * g * t
Substituting the given value for g and the calculated value for t, we have:
v0y = (1/2) * 9.8 m/s^2 * 0.519 s ≈ 2.539 m/s
Therefore, the y-component of the initial velocity required for the salmon to just reach the ledge is approximately 2.539 m/s.
To summarize:
v0x = 2.390 m/s
v0y = 2.539 m/s