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
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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(52810) (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) (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
==================
b. When does the object strike the ground?
h(t) = -4.9t^2 + 17.2t +25
When h(t) = 0
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