the tallest volcano in the solar system I the 23 km tall Martian volcano, Olympus Mons. An astronaut drops a ball off the rim of the crater and that he fee fall acceleration of the ball remains constant throughout he ball's 23 km fall at a value of 3.3 m/^. (We assume that the crater is a deep as the volcano is ta, which is no usually the case in nature.) Find the time for the ball to reach the crater floor? Answer in units of s
To find the time it takes for the ball to reach the crater floor, we can use the kinematic equation:
\(d = \frac{1}{2} \cdot a \cdot t^2\)
Where:
- \(d\) is the distance traveled by the ball (given as 23 km or 23,000 m)
- \(a\) is the acceleration (given as 3.3 m/s^2)
- \(t\) is the time we need to find
Rearranging the equation, we get:
\(t = \sqrt{\frac{2d}{a}}\)
Substituting the given values, we have:
\(t = \sqrt{\frac{2 \cdot 23,000}{3.3}}\)
Calculating this expression, we find:
\(t \approx \sqrt{13,939.39} \approx 118.05\) seconds
Therefore, the time it takes for the ball to reach the crater floor is approximately 118.05 seconds.
To find the time for the ball to reach the crater floor, we can use the equation of motion:
h = (1/2) * g * t^2
Where:
h = height (in this case, 23 km or 23,000 m)
g = acceleration due to gravity (3.3 m/s^2, as mentioned in the question)
t = time (which we want to find)
First, let's convert the height into meters:
h = 23,000 m
Now, let's rearrange the equation to solve for time (t):
2h = g * t^2
t^2 = (2h) / g
t = √[(2h) / g]
Substitute the given values into the equation:
t = √[(2 * 23,000) / 3.3]
Calculating this expression, we find:
t ≈ √(13,939.3939)
Therefore, the time for the ball to reach the crater floor is approximately:
t ≈ 118.03 seconds (rounded to two decimal places)
So, the answer in units of seconds is 118.03 s.