Answer:
The salmon reaches the waterfall with a minimum speed of approximately 20.343 m/s
Step-by-step explanation:
The given parameters are;
The horizontal distance from the waterfall where the salmon starts to jump = 2.00 m
The height of the waterfall, h = 0.55 m
The angle (to the horizontal) at which the salmon jumps, θ = 32°
From
² = 2·g·h, we have;
Where;
g = The acceleration due to gravity = 9.8 m/s²
The minimum vertical velocity required,
is given as follows;
= √(2·g·h) = √(2 × 9.8 m/s² × 0.55 m) = 10.78 m/s
The minimum time, t, it will take the salmon to reach the height of the water fall is given as follows;
= gt
t =
.g = 10.78/9.8 = 1.1 seconds
The vertical velocity
= u × sin(θ)
Therefore, the initial velocity, u =
/sin(θ) = 10.78/(sin(32°)) ≈ 20.343 m/s
The horizontal component of the initial speed = 20.343 m/s × cos(32°) ≈ 17.252 m/s
Therefore, the horizontal distance covered in the 1.1 seconds = 1.1 × 17.252 = 18.9772 meters, which is larger than the 2.00 m distance from the waterfall, therefore, the salmon reaches the waterfall with a minimum speed of 20.343 m/s