SOLUTION: You throw a ball to your friend. The ball leaves your hand 5 feet above the ground and has an initial vertical velocity of 50 feet per second. Your friend catches the ball when it

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: You throw a ball to your friend. The ball leaves your hand 5 feet above the ground and has an initial vertical velocity of 50 feet per second. Your friend catches the ball when it       Log On


   

Question 801192: You throw a ball to your friend. The ball leaves your hand 5 feet above the ground and has an initial vertical velocity of 50 feet per second. Your friend catches the ball when it falls to a height of 3 feet.
A.) Write a function that gives the ball's height (in feet) t seconds after you throw it.
i got: h(t)= -5t^2+50t
B.) How long is the ball in the air?
i set it equal to zero and got 10 seconds
C.) Describe 3 methods you could use to find the maximum height of the ball. Then use all three to find the max. height of the ball.
i only know of 1 way and that is -b/2a
if you could please help me with this it would be greatly appreciated. or if you had any advice it would also give much help. thank you
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
When you use feet and seconds, the 1st term
is +-16t%5E2+ which is the effect of gravity
on the ball. The height above ground is a constant term
because if you set +t+=+0+ you get:
+h%280%29+=+-16%2A0%5E2+%2B+50%2A0+%2B+5+
+h%280%29+=+5+ ft, which it should be
-------------------------------
+h%28t%29+=+-16t%5E2+%2B+50t+%2B+5+
You have to find the 2nd time that +h%28t%29+=+0+
which is when the ball hits the ground.
+h%28t%29+=+-16t%5E2+%2B+50t+%2B+5+
+-16t%5E2+%2B+50t+%2B+5+=+0+
Use the quadratic formula
+t+=+%28+-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29+
+a+=+-16+
+b+=+50+
+c+=+5+
+t+=+%28+-50+%2B-+sqrt%28+50%5E2+-+4%2A%28-16%29%2A5+%29%29+%2F+%282%2A%28-16%29%29+
+t+=+%28+-50+%2B-+sqrt%28+2500+%2B+320+%29%29+%2F+%28+-32+%29+%29+
+t+=+%28+-50+%2B-+sqrt%28+2820+%29%29+%2F+%28+-32+%29+
+t+=+%28+-50+-+53.104+%29+%2F+%28+-32+%29+
+t+=+103.104+%2F+32+
+t+=+3.222+
The ball is in the air 3.222 sec
--------------------------
Here's a plot:
+graph%28+400%2C+400%2C+-1%2C+4%2C+-10%2C+50%2C+-16x%5E2+%2B+50x+%2B+5+%29+
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The t-coordinate of maximum height is at +-b%2F%282a%29++=+-50%2F%28-32%29+
+t%5Bmax%5D+=+1.563+
Plug this back into the equation to get +h%5Bmax%5D+
--------------------
Another way is to find the +t+ value midway between
+h%28t%29+=+0+ and +h%28+3.222%29+=+0+%2C+and+then%0D%0Afind+the+%7B%7B%7B+h+ value
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Another way is just to make the plot