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Sagot :
Answer:
a) 2.85 m/s
b) -24.01 m/s
c) 1
Explanation:
From the question, we can attest that the motion of the fish that was dropped by the seagul is that of a projectile motion. This motion is made up of two other motions which are, a horizontal uniform motion and a vertical motion, at constant acceleration. From the question, we are asked to find the horizontal motion. And then, the horizontal component of the fish's velocity does not change, therefore its the horizontal component of the fish's velocity is 2.85 m/s
To find the vertical component of the fish's velocity, we use the equation
v(y) = u(y) + gt
Where u(y) is the initial velocity which is zero, and g is the acceleration due to gravity which is -9.8 m/s. We are also told from the question that it took 2.45 s, and that's our time t. Applying this into the equation, we have
v(y) = 0 + -9.8 * 2.45
v(y) = -24.01 m/s
For the 3rd part, the horizontal component of the fish's velocity would decrease
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