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.
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
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
Vo + g*Tr = 0
20 + (-10)Tr = 0
Tr = 2 s. = Rise time to reach max. ht.
h = 3 + 20*2 - 5*2^2 =
To deduce the equation of the axis of symmetry, we need to find the horizontal coordinate (t-value) at which the object reaches its highest point.
The equation of the axis of symmetry for a quadratic equation in the form of y = ax^2 + bx + c is given by t = -b/2a.
In this case, the equation for height is h = 3 + 20t - 5t^2. Comparing this to the standard form ax^2 + bx + c, we can see that a = -5 and b = 20.
Using the equation of the axis of symmetry, t = -b/2a, we substitute the values of a and b:
t = -(20) / [2(-5)]
t = -20 / -10
t = 2
Therefore, the equation of the axis of symmetry is t = 2.
To calculate the greatest height the object reaches, we can substitute the value of t = 2 into the equation h = 3 + 20t - 5t^2:
h = 3 + 20(2) - 5(2)^2
h = 3 + 40 - 5(4)
h = 3 + 40 - 20
h = 23
So, the greatest height the object reaches is 23 meters above ground level.