Question

Suppose a chinook salmon needs to jump a waterfall that is 1.43 m high. (a) If the fish starts from a distance 1.08 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.

Answer 1

Answer:

The solution to this question is as follows

Hence the x component of the initial velocity of the Chinook salmon is 2 m/s

The y component of the initial velocity of the salmon to scale the waterfall is 5.3 m/s

see explanation below

Explanation:

To solve this question, we need to list out the known values from where the number of unknowns and the right equation of motion can be decided thus;

Height of the waterfall = 1.43m

Distance of the fish from the base of the ledge over which the water fall flows = 1.08m

Thus the equations of motion required include

v² = u² - 2×g×S and

v = u -g×t

where u is the the initial velocity we have

v = 0 for vertical velocity at the top of the ledge

thus u = gt at the top of the ledge and S = 1.43m

this gives (gt)²-2×1.43×g =0 or

2.86 =gt² or t² = 2.86÷9.81 = 0.29 or t = 0.54s

From where the vertical component of the velocity = 9.81×0.54 = 5.3m/s

similarly the horizontal component can be found by v = distance÷time

= (horizontal distance from the bottom of the ledge) ÷ (the time it takes to scale the ledge) = 1.08÷0.54 = 2 m/s

Hence the x component of the initial velocity of the Chinook salmon is 2 m/s

The y component of the initial velocity of the salmon to scale the waterfall is 5.3 m/s