SOLUTION: A rock is thrown upward at an initial velocity of 30 ft/sec from a height of 20 feet. What will its maximum height be?

Question 101390: A rock is thrown upward at an initial velocity of 30 ft/sec from a height of 20 feet. What will its maximum height be?
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
The equation you are looking for is:
.

.
In this equation, the letters represent the following quantities:
.
represents the height of the rock at t seconds after it is launched.
.
represents the acceleration of an object due to gravity. In English units it is generally
accepted to be 32 ft/sec^2
.
is the initial velocity at which the is launched. In this problem it
will have a positive sign because it is launched upward.
.
is the height above ground from which the is launched. Its sign will be positive
because it is above ground level.
.
is the number of seconds after the rock is launched.
.
With this equation and the definitions of the variables in mind, we can substitute the
given values into the equation. g is 32, the initial velocity is +30 ft/sec, and the initial
height of launch is 20 ft. The equation becomes:
.

.
and you can divide the multiplier of the term. 32 divided by 2 is 16 so the equation
becomes:
.

.
Note that this is a quadratic equation of the form:
.

.
If you are familiar with the quadratic formula you know that when you set y equal to zero
and in standard form the equation becomes:
.

.
and the quadratic formula tells you that the values of x that make this happen are given
by the equation:
.

.
You can find the "peak" of this curve in a couple of ways. It will occur when the two values
of x are found and averaged. The average of the two values of x will be the value of x
where the peak occurs. Then you can substitute this average value of x into the equation
and it will tell you the value of y at the peak of the graph.
.
The average value of x is also equal to which is the first term in the solution
to the quadratic equation.
.
Let's use this second method to find the value of t we are looking for. We'll begin by
comparing our gravity equation to the equation of the quadratic form. That is: compare:
.

.
and
.

.
a is the multiplier of the squared term and in our gravity equation, that multiplier
is -16.
.
b is the multiplier of the x term and in our gravity equation the multiplier of the corresponding
t term is +30.
.
Now we can substitute these values into and we get:
.

.
So at seconds after launch the rock will be at maximum height. (In decimal form
this is 0.9375 seconds after launch.)
.
Now if we return to the height equation and substitute 0.9375 seconds for t we can get the
height at the peak of the path of the rock.
.
Start with:
.

.
Substituting 0.9375 for t makes this equation become:
.

.
Put your calculator to work and you should get:
.

.
So the rock rises to a maximum height of 34.0625 feet.
.
Lots of work to follow here. Hope this helps you to see how you might solve this problem.
.

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