Question 651651: the height h(t) of a projectile measured from ground-level is given by
h(t) = -t^2+8t where t represents the elapsed time in seconds since it was launched
a) can the projectile hit a target located at a height of 20 cm?
b) at what instant does the projectile hit the ground?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 
based on the graph that i just showed you, the projectile will not be able to hit a target at 20 cm.
the formula used is:
h = -t^2 + 8t
note that the graph requires x rather than t, so i replace t with x in the equation for graphing purposes.
note that the graph displays the answers as y, so assume that y = h.
the equation for graphing is:
y = -x^2 + 8x
nothing was changed except the name of the variables used which doesn't have any impact on the relationship expressed by the equation.
i'll work with this form of the equation:
y = -x^2 + 8x
this is a quadratic equation and the min/max point of a quadratic equation is expressed as (x,f(x)) where x is equal to -b/2a.
note that the standard form or a quadratic equation is:
ax^2 + bx + c = 0
if you set y equal to 0, then your equation can be expressed as:
-x^2 + 8x = 0
from the standard form of the equation, we get:
a = -1
b = 8
c = 0
the formula for the max/min point of a quadratic equation tells us that we can find x by using the formula of:
x = -b/2a
substituting 8 for b and -1 for a,this equation becomes:
x = -8/-2 which becomes x = 4
our equation is:
y = -x^2 + 8x which can also be shown as:
f(x) = -x^2 + 8x.
when x = 4, this equation becomes:
f(4) = -(4^2) + 8(4) which simplifies to:
f(4) = -16 + 32 which simplifies to:
f(4) = 16
the max/min point of the quadratic equation is (4,16)
this is a max point as shown on the graph.
it can also be determined by the fact that a is negative.
if a is negative, the quadratic equation points up and opens down and has a max point at x = -b/2a
if a is positive, the quadratic equation points down and opens up and has a min point x = -b/2a
since a is negative, x = -b/2a finds the max point as (once again) shown on the graph.
to find the point at which the projectile returns to the ground, set y = 0 and solve the equation.
the equation to solve (using x rather than t) is:
-x^2 + 8x = 0
factor out the x to get:
x * (-x+8) = 0
this equation will be true if either:
x = 0
or (-x+8) = 0
solve for x in (-x+8) = 0 as follows:
remove the parentheses to get:
-x + 8 = 0
add x to both sides of the equation to get:
8 = x
this is equivalent to:
x = 8
your possible solutions are:
x = 0 or x = 8
x = 0 would be the point at which the projectile is launches.
x = 8 would be the point at which the projectile returns to the ground.
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