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Answer:
The maximum height that the fish can jump is 2.19 m.
Explanation:
Hi there!
Please, see the attached figure for a better understanding of the problem.
The motion of the salmon is a parabolic one because when it jumps, it already has a horizontal velocity (see figure).
The position and velocity vectors of the salmon at a time t, can be calculated as follows:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
v = (v0x, v0y + g · t)
Where:
r = position of the salmon at time t.
x0 = initial horizotal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration of gravity.
Looking at the figure, notice that at the maximum height, the vertical velocity is zero (because the velocity vector is horizontal). Using the equation of the vertical component of the velocity, we can obtain the time at which the salmon is at its maximum height:
vy = v0y + g · t
To find the initial vertical velocity, v0y, let´s look at the figure. Notice that the initial velocity is the hypotenuse of the triangle formed with the horizontal velocity and the vertical velocity. Then:
v0² = v0x² + v0y²
Solving for v0y:
v0y = √(v0² - v0x²)
v0y = √((6.75 m/s)² - (1.65 m/s)²)
v0y = 6.55 m/s
Now, using the equation of the vertical component of the velocity at the maximum height (vy = 0):
vy = v0y + g · t
0 = 6.55 m/s + (-9.8 m/s²) · t
-6.55 m/s / -9.8 m/s² = t
t = 0.67 s
Now, using the equation of the vertical position at t = 0.67 s, we can find the maximum height:
y = y0 + v0y · t + 1/2 · g · t²
y = 0 m + 6.55 m/s · 0.67 s + 1/2 · (-9.8 m/s²) · (0.67 s)²
y = 2.19 m
The maximum height that the fish can jump is 2.19 m.
Answer:
The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
Explanation:
From the question we are told that
The distance between earth and Retah is
Here c is the peed of light with value
The time taken to reach Retah from earth is
The velocity of the spacecraft is mathematically evaluated as
substituting values
The time elapsed in the spacecraft’s frame is mathematically evaluated as
substituting value
=>
So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame