Asked by Ashley

A high-jumper's path on a jump can be represented by the equation h=-16t^2+80t. At what time(s) does the jumper reach 84 ft? Show or explain your work.
Ok, so i have no idea how to do this, does anyone know how to solve this question?
Answered by Damon
Do you know how to solve a quadratic?
16 t^2 - 80 t - 84 = 0
Answered by Damon
Hey --- no way he jumped 84 feet ---- typo
Answered by Ashley
I know, that’s what i thought but i wasn’t sure. And yeah, i just forgot. Thx
Answered by Damon
and no way his initials peed was 80 feet/second
Answered by Ashley
Omg....u answered at 420. That’s beautiful
Answered by Damon
how hi could really go if he could jump at 80 ft/s ?
v = 80 - 32 t
v = 0 at top
t = 80/32 = 2.5 seconds upward (wow)
h = 0 + 80(2.5 ) - 16 (2.5)^2
= 200 - 100
= 100 feet (some kind of a jump)
Answered by Ashley
Yeah...that’s...high.
Answered by Damon
that is 4t^2 - 20 t -21 = 0 by the way
Answered by Damon
https://www.mathsisfun.com/quadratic-equation-solver.html
Answered by Damon
You race 420s ?
Answered by henry2,
h = -16t^2 + 80t = 84 Ft.
Divide both sides by 4:
-4t^2 + 20t = 21.
-4t^2 + 20t - 21 = 0.
Use Quadratic formula to find t:
t = (-B +- sqrt(B^2 - 4AC))/2A,
t = (-20 +- sqrt(400 - 336))/(-8),
t = (-20 +- 8)/-8 = 1.5 s, and 3.5 s.

Answered by henry2,
Well, I guess I am the only human that can jump to a height of 84 ft. LOL.
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