An arrow is shot vertically upward from a platform 48 ft high at a rate of 233 ft per sec. When will the arrow hit the ground? Use the formula: h = −16t2 + v0t + h0.

Answer 1

it hits the ground, naturally, when h=0. So, just find t when

-16t^2 + 233t + 48 = 0

Answer 2

idon't understand
how to perform this operation

Answer 3

Just use the quadratic formula, which you learned in Algebra I:

t =

-233±√(233^2 - 4(-16)(48))
--------------------------------
      2(-16)

t = (233±√57361)/32

or,

t = -0.20 or 14.76

We don't allow negative time, so the height is zero (the arrow is back on the ground) after 14.76 seconds.