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In a soccer match, a ball is kicked from the ground with an initial velocity of 96ft/s².
The function for the projectile motion of the ball is H(t)=-16t²+96t.
When will the ball reach its maximum height? What is the maximum height?
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This quadratic function is
h(t) = -16t^2 + 96t = -16t*(t-6).
It has zeroes (y-intercepts) t= 0 and t= 6.
The maximum of this quadratic function is reached half-way between the roots t= 0 and t= 6,
and this half-way point is t= 3 seconds.
It means that the ball reaches its maximum height at t= 3 seconds, and the maximum height is equal to
h(3) = -16*3^2 + 96*3 = -16*9 + 96*3 = 144.
ANSWER. The ball reaches the maximum height of 144 feet at the time t= 3 seconds.
Solved and thoroughly explained.
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For better understanding, see the lessons
- Introductory lesson 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
Consider these lessons as your textbook, handbook, tutorials and (free of charge) home teacher.
Read them attentively and learn how to solve this type of problems once and for all.
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.