A large boulder is ejected vertically upward from a volcano with an initial speed of 39.0m/s . Air resistance may be ignored.

(a) At what time after being ejected is the boulder moving at a speed 19.0m/s upward?

(b) At what time is it moving at a speed 19.0m/s downward?

a. V = Vo + g*t = 19 m/s.

39 - 9.8*t = 19
-9.8t = 19-39 = -20
t = 2.04 s.

b. V = Vo + g*t = -19 m/s.
39 - 9.8t = -19
-9.8t = -58
t = 5.92 s.

If the question is to determine the maximum height of the boulder, what would the answer be?

To solve this problem, we can use the equations of motion for a vertically launched projectile. The equation we will use is:

v = u + at

where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time

In this case, the initial velocity (u) is 39.0 m/s and the acceleration (a) is -9.8 m/s^2 (taking downward direction as negative). We can solve for time (t) in both cases.

(a) When the boulder is moving upward at a speed of 19.0 m/s:

19.0 = 39.0 - 9.8t

Rearranging the equation:

9.8t = 39.0 - 19.0
9.8t = 20.0

Dividing both sides by 9.8:

t = 20.0 / 9.8
t ≈ 2.04 seconds

Therefore, at approximately 2.04 seconds after being ejected, the boulder is moving at a speed of 19.0 m/s upward.

(b) When the boulder is moving downward at a speed of 19.0 m/s:

-19.0 = 39.0 - 9.8t

Rearranging the equation:

9.8t = 39.0 + 19.0
9.8t = 58.0

Dividing both sides by 9.8:

t = 58.0 / 9.8
t ≈ 5.92 seconds

Therefore, at approximately 5.92 seconds after being ejected, the boulder is moving at a speed of 19.0 m/s downward.

To solve this problem, we can use the equations of motion for an object undergoing constant acceleration, such as the boulder ejected from the volcano. The key equation we will use is:

v = u + at

where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time

(a) To find the time when the boulder is moving at a speed of 19.0 m/s upward, we can set v = 19.0 m/s, u = 39.0 m/s, and solve for t.

19.0 m/s = 39.0 m/s + at

Rearranging the equation to solve for t:

t = (19.0 m/s - 39.0 m/s) / a

Since the boulder is moving upward, the acceleration is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2.

t = (19.0 m/s - 39.0 m/s) / (-9.8 m/s^2)

Calculating:

t = (-20.0 m/s) / (-9.8 m/s^2)

t ≈ 2.04 seconds

Therefore, the boulder will be moving at a speed of 19.0 m/s upward approximately 2.04 seconds after being ejected.

(b) To find the time when the boulder is moving at a speed of 19.0 m/s downward, we need to consider that the initial velocity of the boulder is upward. In this case, the acceleration will also be in the downward direction.

Using the same equation:

v = u + at

Setting v = -19.0 m/s (negative sign indicates moving downward):

-19.0 m/s = 39.0 m/s + at

Solving for t:

t = (-19.0 m/s - 39.0 m/s) / a

Since the acceleration is in the downward direction, it remains -9.8 m/s^2.

t = (-19.0 m/s - 39.0 m/s) / (-9.8 m/s^2)

Calculating:

t = (-58.0 m/s) / (-9.8 m/s^2)

t ≈ 5.92 seconds

Therefore, the boulder will be moving at a speed of 19.0 m/s downward approximately 5.92 seconds after being ejected.