The apparent backward motion of a planet along the celestial sphere is called:
1 answer:
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The final velocity of the rock before it touches the ground is 28 m/s.
Answer:
Explanation:
As the rock is thrown down, this means the acceleration due to gravity will be exerting on the rock. So the rock will be exhibiting a free fall motion. Thus, the acceleration of the rock will be equal to the magnitude of acceleration due to gravity. Then using the third equation of motion, we can determine the final velocity of the rock provided the values for initial velocity, displacement and acceleration is given in the problem itself.
So the acceleration is equal to 9.8 m/s² due to its free fall motion and displacement will be equal to the height of the tower which is given as 30 m. And the initial speed of the rock is stated as 14 m/s. The initial speed is represented as u, final speed is represented as v, displacement is represented as s and acceleration is represented as a.
Then, 2 × 9.8 × 30 = v²-(14)²
v²=784
v= 28 m/s
So the final velocity of the rock before it touches the ground is 28 m/s.
Answer:
Explanation:
As we know that the force equation at the top is given as
now we know that
so we have
so we will have