Question 214728
 An arrow is shot vertically upward. Three seconds later, it is at a height of 35m.
What was the initial speed of the arrow?
:
Using the equation -4.9t^2 + vt + c = H
Where
h = height after t seconds (35 meters) 
t = time in seconds (3)
-4.9t^2 = the downward pull of gravity
v = upward velocity
c = initial height (0 in this problem)
:
-4.9(3)^2 + 3v = 35
-4.9(9) + 3v = 35
-44.1 + 3v = 35
3v = 35 + 44.1
3v = 75.1
v = {{{75.1/3}}}
v = 26.37 m/sec initial velocity
:
How long is the arrow in flight from launch to returning to the initial height? (Assume the arrow falls vertically downward.)
-4.9t^2 + 26.37t = 0
Factor out -4.9t
-4.9t(t - 5.38) = 0
Two solutions
t = 0 sec; arrow on it's way
t = 5.38 sec when it returns to earth (time of flight)