Question

A cannonball is fired into the air. Its height, h, in metres, after t seconds is given by the relation h = –4.9(t – 3) 2 + 46.1 a) What is the vertex, and what does it represent in this situation? b) From what height was the cannonball fired? (hint: t = 0) c) After how many seconds (rounded to first decimal) will the cannonball strike the ground? (hint: h=0)

Answer 1

Step-by-step explanation:

a) Hey, you know what h is when t = zero!

b ) h = -4(1) + 16(1) + 9

= 21

c) If you do not know any calculus, which I assume you do not or you would not be asking, then you must complete the square to find the vertex of the parabola

t^2 - 4 t = -h/4 + 9/4

t^2 - 4 t + 4 = -h/4 + 9/4 + 16/4

(t - 2)^2 = -(1/4) (h - 25)

so

max height = 25 at t = 2

d ) 0 =-4t^2+16t+9

or

t^2 - 4 t - 9/4 = 0

t = [ 4 +/- sqrt(16 -81/4) ] /2

no real solution

The reason is that your original equation is in error

h=-4t^2+16t+9.

It should be assuming feet and seconds and g = -32 ft/s^2

h = (1/2) g t^2 + Vi t + Hi

h = -16 t^2 + Vi t + Hi