using the quadratic formula for projectile motion

Question

if H= 0= t²-(2V₀/g)t - (2h₀/g) and we want to solve for t,

how is the quadratic formula for the projectile t1= V_o/g + - sqrt ( (Vo/g)² + 2ho/g )

I'm confused about where the 2 on b went inside the square root. Does it cancel out because of the 2 in the denominator.... if that's the case how does the 2 remain with Ho/g?

Answer 1

h(t) = t^2 -2vot/g - 2ho/g = 0

mutiply by -g/2 to get

h(t) = -gt^2/2 +vot +ho = 0 which is the more standard form

to find the tiime t when it hits the ground

ho = initial height at time t=0

vo = initial velocity at time t=0

-g = deceleration due to gravity

g/2 = usually either 16 feet or 4.9 if measured in meters

t=-b/2a + or - (1/2a)sqr(b^2 -4ac)

where a= -g/2, b= vo, c=ho

t =-vo/g + or - (1/g)sqr(vo^2 -4gho)

Answer 2

Answer 3

The discriminat shall have a 4 that can be factored out of the radical to become 2 this with the 2 in 2v₀/g reduces with the 2 in the denominator.