SOLUTION: The height of an object above the ground at time t is given by s(t) = -0.5gt^2 +ut where u is the initial velocity(constant ) upward (m/s) and g is the constant acceleration due

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Question 1094255: The height of an object above the ground at time t is given by
s(t) = -0.5gt^2 +ut
where u is the initial velocity(constant ) upward (m/s) and g is the constant acceleration due to gravity.
1. At whAT HEIGHT IS THE OBJECT INITIALLY ?
2. for how long is the object in the air before it hits the ground ?
3. when will the object reach its maximum height?
4. what is this maximum height ?
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
For the first question, the answer is: initially the object was at the ground level.

The initial height was 0.


For other questions, see 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.


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.