A Chinook salmon can jump out of water with a speed of 7.20 m/s . How far horizontally can a Chinook salmon travel through the air if it leaves the water with an initial angle of =29.0° with respect to the horizontal? (Neglect any effects due to air resistance.) d=

Explanation:

The vertical component of the salmon's velocity is 7.2 m/s x sin 29 = 3.49 m/s

If g = 9.81 m/s^2, the salmon takes

(3.49 m/s) / (9.81 m/s^2) = 0.356 s to reach the highest point of its trajectory.

It takes another 0.356 s to fall back into the water again.

So the salmon is out of the water for a total of 0.712 s.

In this time the salmon travels horizontally with a velocity of 7.2 m/s x cos 29 = 6.30 m/s

We can now calculate the horizontal distance tavelled by multiplying the horizontal velocity by the time spent out of water;

0.712 s x 6.30 m/s = 4.48 m

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f=ma newton

f=100x2.5=250n