SOLUTION: A skateboarder decides to jump off a ramp. The path of the jump from the ramp can be approximated by: h=-3t^2+6t+1, where h is the height above the ground in metres, s is the horiz

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A skateboarder decides to jump off a ramp. The path of the jump from the ramp can be approximated by: h=-3t^2+6t+1, where h is the height above the ground in metres, s is the horiz      Log On


   

Question 1137550: A skateboarder decides to jump off a ramp. The path of the jump from the ramp can be approximated by: h=-3t^2+6t+1, where h is the height above the ground in metres, s is the horizontal displacement and t is the time after leaving the ramp in seconds. Show that the maximum height reached by the skateboarder is 4m.
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
the function is a convex upward parabola.
the t value at the vertex, the highest point, is at -b/2a=-(6)/-6, or 1
f(1)=-3+6+1=4 meters.
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C-3x%5E2%2B6x%2B1%29

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

            I'd like to make couple of notices,  regarding this post and the problem formulated in it. 


1.  If the events, described in the post, take place at the Earth surface (and not at the moon surface, for example),

    then the coefficient at t^2 must be equal to -5 (or -4.9), which is half of the gravity acceleration

    at the Earth surface, expressed in m/s^2.  It can not be equal to -3.



2.  The condition declares that "s is the horizontal displacement", but  " s "  absents in the post.

    So, the given part of the post describes the vertical movement only, without touching its horizontal displacement.



3.  I make these my notes to help to the person, who created this problem, to get understanding that his (or her) formulation  

    is partly wrong and partly incomplete, and to help to him (or to her) to develop further his (or her) knowledge on the subject.


Happy learning !

===================

To wider your horizon and to better learn the subject,  see my lessons
    - Problem on a projectile moving vertically up and down
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower
    - Problem on a toy rocket launched vertically up from a tall platform
in this site.

Consider these lessons as your textbook,  handbook,  tutorials and  (free of charge)  home teacher.
Read them attentively and learn how to solve this type of problems once and for all.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic "Projectiles launched/thrown and moving vertically up and dawn".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.