Question 72462This question is from textbook
: A body is thrown straight from the ground with an initial velocity of 96 feet per second. Neglecting air resistance, its height h, measured in feet, after t seconds, is given by the formula:
h=96t-16t(squared)
A)After how many seconds does the body reach its maximum height?
B)What is the maximum height?
This question is from textbook
Answer by funmath(2933)   (Show Source):
You can put this solution on YOUR website! A body is thrown straight from the ground with an initial velocity of 96 feet per second. Neglecting air resistance, its height h, measured in feet, after t seconds, is given by the formula:
h=96t-16t(squared)
A)After how many seconds does the body reach its maximum height?
B)What is the maximum height?
:
If you're supposed to use calculus, let me know and I'll redo this.
A) There is two ways to approach this, first you can realize that the body will reach its peak half way between the time it is thrown and the time it lands.
16t=0 or 6-t=0
16t/16=0/16 or 6-6-t=0-6
t=0 or -t=-6
t=0 or t=6
The body is thrown at t=0 s and lands at t=6 s, halfway between those two points in time is 1/2(6)=3 s.
Another way to do this is to find the vertex of the parabola given the equation.
There's a formula for that, or you can complete the square. If you're supposed to complete the square, let me know and I'll redo this.
The formula is  for a quadratic equation written in standard form  .
a=-16 b=96 and c=0
t=3 s.
:
B)
To find the maximum height, substitute 3 in for t and solve for height.
h(3)=288-144
h(3)=144 ft.
Happy Calculating!!!!
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