SOLUTION: A skateboarder jumps off a ledge. His path is modeled by the relation h= -0.8d^2 + 0.8d + 1.6, where h is his height above the ground and d is his horizontal distance from the led

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A skateboarder jumps off a ledge. His path is modeled by the relation h= -0.8d^2 + 0.8d + 1.6, where h is his height above the ground and d is his horizontal distance from the led      Log On


   

Question 1197659: A skateboarder jumps off a ledge. His path is modeled by the relation
h= -0.8d^2 + 0.8d + 1.6, where h is his height above the ground and d is his horizontal distance from the ledge, both in metres.
a) What is the height of the ledge?
b) Factor to find the zeros of the relation.
c) At what point will the skateboarder land on the ground?
d) Graph the relation.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi  
A skateboarder jumps off a ledge. 
His path is modeled by the relation:
h= -0.8d^2 + 0.8d + 1.6
Graph:
 .
a) What is the height of the ledge? 1.6m
b) Factor to find the zeros of the relation. d = 2 is the appropriate zero
c) At what point will the skateboarder land on the ground?
h = 0, Graph tells us: 
the skateboarder will land on the ground  2m from the ledge

Graph also tells us the max height jumped is 1.8m at .5m from ledge.
 Am using a FREE graphing software downloaded from  Padowan.dk
Wish You the Best in your Studies.