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An eagle is flying horizontally at a speed of 4.10 m/s when the fish in her talons wiggles loose and falls into the lake 6.20 m below. Calculate the velocity (in m/s) of the fish relative to the water when it hits the water. (Assume that the eagle is flying in the +x-direction and that the +y-direction is up.)

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obasola1

Answer:

V=11.74m/s, 69.59°

Explanation:

From newtons equation of motion ,we know that

V^2 = u^2+2gh

for the rt

For the vertical component of the speed

Vy^2=V0^2+ 2gh

Vy= the final speed in the vertical axis

V0= initial speed, 0m/s

g= acceleration due to gravity 9.81m/s

h=height/distance between the eagle and the lake

Vy^2= 2*9.81*6.2

Vy =√121.644

Vy=11.02m/s

The resultant speed will be

V=(Vy^2+Vx^2)^0.5

V=(11.02^2+4.1^2)^0.5

V=137.7241^0.5

V=11.74m/s

Direction

Tan^-1(11.02/4.1)

β=69.59°

V=11.74m/s, 69.59°

Answer =11.74m/s, b°

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