Asked by diane
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.
Answers
Answered by
Steve
it hits the ground, naturally, when h=0. So, just find t when
-16t^2 + 233t + 48 = 0
-16t^2 + 233t + 48 = 0
Answered by
Anonymous
idon't understand
how to perform this operation
how to perform this operation
Answered by
Steve
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.
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.
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