the general formula is:

y(t) = 1/2at^2 + V(0)t + y(0) (a is constant and it’s equal to g=9.81 m/(s^2) as it’s a free fall)

but with the formula you gave what is s?? if we consider the units, s must be a height. I’ll suppose that it’s the initial height (15ft).

we’re looking for t when h=0

h = -16t^2 + v(0)t + s

0 = -16t^2 + 5t + 15

using quadratic formulas, you can find 2 values of t, but just 1 positive value. as time cannot be negative, the positive value is the correct one, and it’s the one found by calculating (i cant put square root so i’ll put sr):

(-5 – sr(25-4*(-16)*15))/(2*-16) = mmmMMMMMMM u can do it…

it’s almost T = 1.125s

supply up and picture approximately each and each achieveable answer: a) Acceleration is 0? – it is real. on the apex of the trajectory the bullet could have 0 acceleration, as long as we don’t evaluate deceleration through air resistance. b) speed is 0? – fake. the only way this could be real is that if it have been on the floor or if it hit some thing. Now, the vertical speed could be 0 however the forward (horizontal) isn’t. c) top is 0? – needless to say no longer considering that we did shoot up and not point. d) All of above? – won’t be able to be considering that we proved b and c to be fake e) None? – no longer this the two seeing as we chanced on the acceleration to be 0.

H= -16t^2 + 5t + 15

if you think about it, H = y and t = x

it’s a simple quadratic graphing problem

Your answer is:

T = 1.137 seconds about.

plug in 5 for v and 15 for h

then solve for t

Respect Beer.