SOLUTION: An object is launched at 17.2 m/s from a 25-meter tall platform. The equation for the object's height h at time t seconds after launch is: h(t) = -4.9t^2 + 17.2t +25, where h is in

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: An object is launched at 17.2 m/s from a 25-meter tall platform. The equation for the object's height h at time t seconds after launch is: h(t) = -4.9t^2 + 17.2t +25, where h is in      Log On


   

Question 1077426: An object is launched at 17.2 m/s from a 25-meter tall platform. The equation for the object's height h at time t seconds after launch is: h(t) = -4.9t^2 + 17.2t +25, where h is in meters.
a. Graph this equation.
b. When does the object strike the ground?
Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52805) About Me  (Show Source):
You can put this solution on YOUR website!
.
Read the lessons
    - Problem on a projectile moving vertically up and down
    - Problem on a toy rocket launched vertically up from a tall platform
in this site.

They contain the solutions to similar problems with detailed explanations.


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".


Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
An object is launched at 17.2 m/s from a 25-meter tall platform. The equation for the object's height h at time t seconds after launch is: h(t) = -4.9t^2 + 17.2t +25, where h is in meters.
a. Graph this equation.
==============
It's a parabola.
Pick values for t and find h.
Plot the points.
Draw a curve thru them
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b. When does the object strike the ground?
h(t) = -4.9t^2 + 17.2t +25
When h(t) = 0