SOLUTION: A model rocket is launched with an initial velocity of 250 ft per second. The height h, in feet, of the rocket t seconds after the launch is given by h = −16t2 + 250t. Ho

Question 1124707: A model rocket is launched with an initial velocity of 250 ft per second. The height h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 250t.
How many seconds after the launch will the rocket be 400 ft above the ground? Round to the nearest hundredth of a second.
Thank you!!!

Answer by ikleyn(52788) (Show Source): You can put this solution on YOUR website!
.

To answer the question, you need to solve the equation


    h(t) = 400,


or, which is the same, the equation


    -16t^2 + 250t = 400.

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To enhance your horizon, there is a bunch of lessons in this site on a projectile thrown/shot/launched vertically up
    - 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

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.

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SOLUTION: A model rocket is launched with an initial velocity of 250 ft per second. The height h, in feet, of the rocket t seconds after the launch is given by h = &#8722;16t2 + 250t. Ho

SOLUTION: A model rocket is launched with an initial velocity of 250 ft per second. The height h, in feet, of the rocket t seconds after the launch is given by h = −16t2 + 250t. Ho