A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou
nd. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.
1 answer:
Answer:
At the highest point the velocity is zero, the acceleration is directed downward.
Explanation:
This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.
I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.
At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
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Velocity is defined by rate of change in the position
which we can also write as
while acceleration is defined as rate of change in velocity
so acceleration and velocity both are rate of change in position and rate of change in velocity with respect to time respectively
out of all above statement the correct answer must be
<u>Acceleration equals change in velocity divided by time. </u>
Answer:
Explanation:
Given:
- mass of rocket,
- time of observation,
- mass lost by the rocket by expulsion of air,
- velocity of air,
<u>Now the momentum of air will be equal to the momentum of rocket in the opposite direction: </u>(Using the theory of elastic collision)