Question

An object is launched at 19.6 m/s from a height of 58.8 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.9t2 +19.6t + 58.8. What is the object's maximum height?

Answer 1

find the vertex of the parabola (or use calculus)

Answer 2

man its been long since ive done that...would you mind giving some clues or somthn?

Answer 3

is this calculus?

Answer 4

supose to be algebra

Answer 5

the vertex occurs at x= -b/(2a)

in y= ax^2 +bx +c

Answer 6

match the letters to the numbers in your equation

Answer 7

once you find x (as a number), use it to find the height by replacing x with the number in the equation

Answer 8

I mean t in this case, rather than x

Answer 9

so 58.8=-4.9(19.6)^2+19.6(19.6) + 58.8 ?

Answer 10

h(t) = -4.9t2 +19.6t + 58.8

this is a parabola (I shot an arrow into the air, it follows a parabola)
the peak occurs at t= -b/(2*a)
where a and b are the coefficients of the generic parabola f(t)= a*t^2 +b*t +c

match up these equations
h(t) = -4.9t2 +19.6t + 58.8
f(t) = a*t^2 +b*t + c
to find a and b. then use them in
t= -b/(2*a)

once you find t as a number, use it in
h(t) = -4.9t2 +19.6t + 58.8
to find the height at that t

Answer 11

thanks @phi