Question

please help me i need to have this by monday!!!!! an arrow is shot straight up in the air with an initial velocity of 70.6 m/s. determine the time of flight, the peak height of the arrow, and the final velocity just befor the arrow hits the ground. i also need to know what the graph of this looks like.

Answer 1

you can use the equation
y = 1/2*g*t^2 + vo*t+ y0

Answer 2

y = 1/2 (-9.8) t^2 + 70.6*t

Answer 3

the time of flight , solve y = 0

0 = 1/2 (-9.8) t^2 + 70.6*t

Answer 4

t = 0
t = 706/49 ~~ 14.408

Answer 5

so it hits the ground around 14.408 seconds

Answer 6

Time to get to max height:
V = Vo + at but V at top = 0 so -Vo= -at
so you can solve this to get time to get to max height.
Ignoring air resistance, tehn the total time of flight by symmetry must be twice what you just calculated
Maximum height
y = yo + Vot + 1/2 at^2 but yo = 0 so y = Vot + 1/2at^2
Final velocity just before it hits the ground
V = Vo + at but Vo at top is zero (it stops moving at the top) so
V = at note: this is half the total time of flight

Which graph do you need?

Answer 7

@perl .. 14.408 is the time for the whole flight or just half

Answer 8

The way @perl solved it, that would be the total time.