**A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0m/s. Air resistance may be ignored. a)At what time after being ejected is the boulder moving at 20.0m/s upward? b)At what time is it moving at 20m/s downward? c)When is the displacement of the boulder from its initial position zero? d) when is the velocity of the boulder zero?**

Answer:

(A)For the first part, we can use the equation :

v=u+at, where v= final velocity, u = initial velocity, a= acceleration, t=time taken

20=40-10*t , where a=10= gravitational acceleration

Time taken=2 seconds

(B) Using the above same equation and proper sign convention with downward as negative and upward as positive,

-20=40-10*t,

we get t=6seconds.

© We can use the equation s=ut+(1/2)at^2

s=40*6-10*(1/2)*6*6=240-180=60m

(D) Again, using the equation, v=u+at

0=40-10*t

t=4 seconds