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The equation is
h(t) = .
Here t is the time after punting (in seconds); h(t) is the height above the ground, in feet.
I assume that the initial velocity 47 ft/s is directed vertically up, although the condition is silent about it.
In this equation:
- the coefficient -16 at t^2 is the (negative) half of the gravity acceleration at the Earth surface, expressed in ft/s^2;
- 47 is the initial velocity, in ft/s; and
- 3 is initial height, in feet.
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To see many solved problems, associated with this equation, look into the 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
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.