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vitfil [10]
2 years ago
7

A chinook salmon needs to jump a waterfall that is 1.5 m high. If the fish starts from a distance of 1.00 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 reach the ledge at the top of its trajectory. Can the fish make this jump? ( A Chinook salmon can jump out of the water with a speed of 6.26m/s)
Physics
1 answer:
____ [38]2 years ago
8 0
 <span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal 

it can cross and reach top of trajectory if its top height h = 1.5m 
and horizontal distance d = (1/2) Range 
--------------------------------------... 
let t be top height time 
at top height, vertical component of its velocity =0 
vy = 0 = u sin p - gt 
t = u sin p/g 
h = [u sin p]*t - 0.5 g[t[^2 
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g 
u^2 sin^2 p/2g = 1.5 
u^2 sin^2 p = 1.5*2*9.8 = 29.4 
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component 
===================== 
t = HALF the time of flight 
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g] 
1 = u^2 sin p cos p/g 
u sin p * u cos p = 9.8 
5.42 * u cos p = 9.8 
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component 
check>> 
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s 
u < less than fish's potential jump speed 6.26 m/s 

so it will able to cross</span>
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Answer:

1. 3 m

2. 27 s

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v₀ = 33 m/s

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Explanation:

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P_o=1.013\times 10^5\ Pa

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We use the equation of motion for vertical component,

s_{y} = u_{y} t+\frac{1}{2} gt^2.

Here, s_{y}   is displacement of bullet, u_{y}  is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

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