The tallest volcano in the solar system is the
17 km tall Martian volcano, Olympus Mons.
An astronaut drops a ball off the rim of the
crater and that the free fall acceleration of the
ball remains constant throughout the ball’s
17 km fall at a value of 3.3 m/s
2
. (We assume
that the crater is as deep as the volcano is tall,
which is not usually the case in nature.)
Find the time for the ball to reach the crater
floor.
Answer in units of s
002 (part 2 of 2) 10.0 points
Find the magnitude of the velocity with which
the ball hits the crater floor.
Answer in units of m/s
H = .5 g t^2
1.7x10^4 = .5 * 3.3 * t^2
1.03x10^4 = t^2
2) multiply the fall time (t) by the acceleration (3.3 m/s^2) to find the final velocity
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 (17 km)
g = acceleration due to gravity (3.3 m/s^2)
t = time
Rearranging the equation to solve for t, we get:
t = sqrt(2h/g)
Substituting the given values:
t = sqrt(2 * 17000 km * 1000 m/km / 3.3 m/s^2)
Calculating this expression, we find:
t ≈ 108.83 seconds (rounded to two decimal places)
So, the time for the ball to reach the crater floor is approximately 108.83 seconds.
To find the magnitude of the velocity with which the ball hits the crater floor, we can use the equation of motion:
v = g * t
Substituting the given values:
v = 3.3 m/s^2 * 108.83 s
Calculating this expression, we find:
v ≈ 358.44 m/s (rounded to two decimal places)
So, the magnitude of the velocity with which the ball hits the crater floor is approximately 358.44 m/s.
To find the time for the ball to reach the crater floor, we can use the kinematic equation:
h = (1/2)gt^2
where h is the height, g is the acceleration due to gravity, and t is the time.
In this case, the height h is equal to 17 km (which we need to convert to meters) and the acceleration due to gravity g is given as 3.3 m/s^2.
Converting 17 km to meters, we get:
17 km = 17,000 m
Now we can plug in the values into the equation and solve for t:
17,000 m = (1/2)(3.3 m/s^2)t^2
Simplifying the equation, we have:
34,000 = 3.3t^2
Dividing both sides by 3.3, we get:
t^2 = 10,303.03
Taking the square root of both sides:
t ≈ 101.5 seconds (approximately)
So, the time for the ball to reach the crater floor is approximately 101.5 seconds.
To find the magnitude of the velocity with which the ball hits the crater floor, we can use another kinematic equation:
v = gt
where v is the velocity and g is the acceleration due to gravity.
Plugging in the values, we have:
v = (3.3 m/s^2)(101.5 s)
Calculating, we get:
v ≈ 334.95 m/s (approximately)
So, the magnitude of the velocity with which the ball hits the crater floor is approximately 334.95 m/s.