The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>
The motion of the boulder is a uniformly accelerated motion, with constant acceleration
a = g = -9.8 
downward (acceleration due to gravity).
By using Suvat equation:
v = u + at
where: v is the velocity at time t
u = 40.0 m/s is the initial velocity
a = g = -9.8
is the acceleration
To find the time t at which the velocity is v = 20.7 m/s
Therefore,

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
The complete question is:
A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?
To learn more about uniformly accelerated motion refer to:
brainly.com/question/14669575
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Answer:
The length of the trail = 22796 ft
Explanation:
From the ΔABC
AC = length of the trail = x
AB = 6100 - 600 = 5500 ft
Angle of inclination
= 15°



x = 22796 ft
Since x = AC = Length of the trail.
Therefore the length of the trail = 22796 ft
Answer:
This question is incomplete
Explanation:
This question is incomplete because the telescope's focal length was not provided. The formula to be used here is
Magnification = telescope's focal length/eyepiece's focal length
The eyepiece's focal length was provided in the question as 0.38 m.
NOTE: Magnification can be described as the power of an instrument (in this case telescope) to enlarge an object. It has no unit and thus the two focal lengths mentioned in the formula above must be in the same unit (preferably meters since one of them is in meters already).