Question

From a platform an object is projected vertically into air. Its height, h metres above ground level after t seconds is modelled by equation h=3+20t-5t^2.
Deduce the equation of axis of symmetry
Calculate the greatest height the object reaches.

Answer 1

h=3+20t-5t^2
well if you know calculus then
dh/dt = 0 at the top
0 = 20 -10 t
t = 2 at top, symmetric about there
if t = 2 then
h = 3 + 20*2 - 5 * 4
= 3 + 40 - 20

Answer 2

if no calculus then complete the square for parabola vertex
h=3+20t-5t^2
5t^2 - 20 t = -h + 3
t^2 - 4 t = (1/5)(-h+3)
t^2 - 4 t + 2^2 = (1/5)(-h+3) + 4
(t-2)^2 = (1/5) (-h+3 +20)
(t-2)^2 = -(1/5)(h-23) there is your vertex

Answer 3

Vo + g*Tr = 0
20 + (-10)Tr = 0
Tr = 2 s. = Rise time to reach max. ht.
h = 3 + 20*2 - 5*2^2 =