SOLUTION: Problem: A ball is thrown vertically upward into the air. If the ball started from a height of 101 feet off the ground, use the following formula to represent the projection of th

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Question 1158667: Problem: A ball is thrown vertically upward into the air. If the ball started from a height of 101 feet off the ground, use the following formula to represent the projection of the ball
H=-16t^2+38t+101
where H is the height of the ball after t seconds have passed. Based on this information, answer the following questions.

1)When does the ball reach its maximum height?
Your answer must be expressed as a decimal rounded to 2 decimal places with correct units. Hint: The maximum occurs at the vertex of the parabola.


2)What is the maximum height of the ball?
Your answer must be expressed as a decimal rounded to 2 decimal places with correct units.


3)When does the ball return to the ground?
You must use the quadratic formula to solve this problem. Your answer must be expressed as a decimal rounded to 2 decimal places with correct units.
Hint: The ball reaches the ground when the height equals zero.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

H=-16t%5E2%2B38t%2B101
1) maximum
you have a parabola, upside down, and max is at vertex
so, write your equation in vertex form
H=%28-16t%5E2%2B38t%29%2B101.......factor out -16
H=-16%28t%5E2-38t%2F16%29%2B101
H=-16%28t%5E2-19t%2F8%29%2B101
H=-16%28t%5E2-%2819%2F8%29%2Bb%5E2%29-%28-16%29b%5E2%2B101.........b=%2819%2F8%29%2F2=19%2F16
H=-16%28t%5E2-19t%2F8%2B%2819%2F16%29%5E2%29%2B16%2819%2F16%29%5E2%2B101
H=-16%28t-19%2F16%29%5E2%2B361%2F16%2B101
H=-16%28t-19%2F16%29%5E2%2B1977%2F16

=> vertex is at(+19%2F16, 1977%2F16)=( 1.19, 123.56)

so, the ball will reach max in t=1.19 seconds

2).
the max height of the ball is 123.56ft

3.
the ball will return to the ground when H=0
0=-16t%5E2%2B38t%2B101..........solve for t using quadratic formula

t=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a

since a=-16,+b=38, and c=101 we have

t=%28-38%2B-sqrt%2838%5E2-4%28-16%29%2A101%29%29%2F%282%28-16%29%29

t=%28-38%2B-sqrt%281444%2B6464%29%29%2F%28-32%29

t=%28-38%2B-sqrt%287908%29%29%2F%28-32%29

t=%28-38%2B-88.93%29%2F%28-32%29

solutions:
t=%28-38%2B88.93%29%2F%28-32%29=-1.59
or
t=%28-38-88.93%29%2F%28-32%29=3.97

disregard negative solution
so, the ball will return to the ground in 3.97 seconds