Hi, I need help to understand this problem solving. Could anybody help me please.

Question

A construction worker is repairing a window tosses a tool to his partner across the street. The height of the tool above the ground is modelled by the quadratic relation h=-5t² + 20t +25, where h is the height in meters and t is the time in seconds after the toss.

a.) How high above the ground is in the window?

b.) If his partner misses the tool, when will it hit the ground?

Answer 1

a) the window is 25 meters above ground

= the constant term in the quadratic polynomial: h(t) = -(a/2)t^2 + vot + ho

ho is the height of the window

general quadractic is

h(t) = -(d/2)t^2 + vot + ho

where ho = the initial height,

vo= initial vertical velocity,

d = deceleration due to gravity = 9.8 m/sec/sec.

Apparently they rounded off the 9.8/2 = 4.9 to 5 m/s^2

a closer approximation is a = 4.90333= 9.80666/2

ho=25 meters, vo = 20 m/sec

initial height = 25 meters

b) set the polynomial = 0 and solve for t

h(t) = -5t^2 + 20t + 25 = 0

5t^2 -20t -25 = 0

t^2 -4t -5 = 0

factor

(t-5)(t+1) = 0

t-5=0, t+1=0

set each factor =0, but ignore the negative t=-1 solution

t = 5 seconds to hit the ground, to reach height of zero,

check the solution

h(5) = -5(5)^2 + 20(5) + 25

= -125 + 100 + 25 = 0

interestingly, this equation only gives the relation between time and height independent of how far the tool was tossed, whether a meter or 2 or 2 kilometers to the other guy.