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Nano-Passion
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Simple Kinematics Problem -- Is the book wrong?
Is the book wrong or do I have faulty logic?
Here is the problem:
An astronaut on a distant planet wants to detrmine its acceleration due to gravity. The astreonaut throws a rock straight up with a velocity of +15m/s and measures a time of 20s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet?
t=(v-u)/a
The solutionn for this problem is 1.5m/s^2. This solution is assumed on the premise that the time going up is the same as the time taken going down. So that time during ascent is 1/2t. Therefore it is 10s.
But that strikes me as rather unintuitive. If you look at the equation t=(v-u)/a, the time taken during ascent would be different than time taken during descent because the rock is thrown with a initial velocity (u) of 15m/s
Homework Statement
Is the book wrong or do I have faulty logic?
Here is the problem:
An astronaut on a distant planet wants to detrmine its acceleration due to gravity. The astreonaut throws a rock straight up with a velocity of +15m/s and measures a time of 20s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet?
Homework Equations
t=(v-u)/a
The Attempt at a Solution
The solutionn for this problem is 1.5m/s^2. This solution is assumed on the premise that the time going up is the same as the time taken going down. So that time during ascent is 1/2t. Therefore it is 10s.
But that strikes me as rather unintuitive. If you look at the equation t=(v-u)/a, the time taken during ascent would be different than time taken during descent because the rock is thrown with a initial velocity (u) of 15m/s