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Let the salmon jump with speed v at an angle of 33.3 degree
So here we can find out the components of his velocity in x and y direction
![v_x = v cos33.2](https://tex.z-dn.net/?f=v_x%20%3D%20v%20cos33.2)
![v_y = v sin33.2](https://tex.z-dn.net/?f=v_y%20%3D%20v%20sin33.2)
now the horizontal displacement of the salmon is 3.15 m so he cover this horizontal distance with constant speed as there is no acceleration in x direction
here we can say
![v_x * t = 3.15](https://tex.z-dn.net/?f=v_x%20%2A%20t%20%3D%203.15)
![v cos33.2 * t = 3.15](https://tex.z-dn.net/?f=v%20cos33.2%20%2A%20t%20%3D%203.15)
![v*t = 3.76](https://tex.z-dn.net/?f=v%2At%20%3D%203.76)
now in Y direction it is an accelerated motion as it is accelerated due to gravity
![y = v_y * t + \frac{1}{2}at^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20%2A%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E2)
![-0.45 = v sin33.2* t - \frac{1}{2}*9.8* t^2](https://tex.z-dn.net/?f=-0.45%20%3D%20v%20sin33.2%2A%20t%20-%20%5Cfrac%7B1%7D%7B2%7D%2A9.8%2A%20t%5E2)
now we will plug in v*t = 3.76
![-0.45 = 3.76 * sin33.2 - 4.9 * t^2](https://tex.z-dn.net/?f=-0.45%20%3D%203.76%20%2A%20sin33.2%20-%204.9%20%2A%20t%5E2)
![4.9 t^2 = 2.51](https://tex.z-dn.net/?f=4.9%20t%5E2%20%3D%202.51)
![t^2 = 0.512](https://tex.z-dn.net/?f=t%5E2%20%3D%200.512)
![t = 0.72 s](https://tex.z-dn.net/?f=t%20%3D%200.72%20s)
now speed v is given by equation above
![v*t = 3.76](https://tex.z-dn.net/?f=v%2At%20%3D%203.76)
![v*0.72 = 3.76](https://tex.z-dn.net/?f=v%2A0.72%20%3D%203.76)
![v = 5.25 m/s](https://tex.z-dn.net/?f=v%20%3D%205.25%20m%2Fs)
so here he must have to jump with minimum speed of 5.25 m/s
Power consumed = P = 78.0 W
Time taken = t = 10 minutes
Energy consumed = E = ?
The power, energy and time are related by the following equation:
P = E / t
Using the values, we get
78 = E/t
So,
E = 780
This means, 780 Joules of energy was used during the 10 minutes
it is Saturn it has the biggest ring
The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was
<em>a</em> = ∆<em>v</em> / ∆<em>t</em> = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²
By Newton's second law, the magnitude of the force <em>F</em> is proportional to the acceleration <em>a</em> according to
<em>F</em> = <em>m a</em>
where <em>m</em> is the object's mass. Solving for <em>m</em> gives
<em>m</em> = <em>F</em> / <em>a</em> = (10 N) / (3.52941 m/s²) ≈ 2.8 kg