SOLUTION: H= -5t^2+20t models the height of a ball that is kicked into the air, where H is the ball's height in meters and t is time in seconds after being kicked. a) How long is the ball

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: H= -5t^2+20t models the height of a ball that is kicked into the air, where H is the ball's height in meters and t is time in seconds after being kicked. a) How long is the ball      Log On


   

Question 1169625: H= -5t^2+20t models the height of a ball that is kicked into the air, where H is the ball's height in meters and t is time in seconds after being kicked.
a) How long is the ball in the air for?


b) When will the ball reach its maximum height?


c) What is the maximum height reached by the ball?


d) At what time(s) will the ball be 15 meters in the air?

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
a) How long is the ball in the air for?
H=+-5t%5E2%2B20t-> t is time in seconds, so set H=0 and solve for t
0=+-5t%28t-4%29
if 0=+-5t+->t=0
or
if 0=+t-4->t=4-> your solution is: the ball is in the air for t=4 seconds
b) When will the ball reach its maximum height?
trajectory is a parabola and maximum is vertex
write H=+-5t%5E2%2B20t in vertex form, complete square
H=+-5t%5E2%2B20t...first factor out -5
H=+-5%28t%5E2-4t%29... complete square
H=+-5%28t%5E2-4t%2Bb%5E2%29-%28-5b%5E2%29...b=4%2F2=2
H=+-5%28t%5E2-5t%2B2%5E2%29%2B5%282%29%5E2
H=+-5%28t-2%29%5E2%2B5%284%29
H=+-5%28t-2%29%5E2%2B20
vertex is at (20, 2)=> the ball reach its maximum in 20 seconds
c) What is the maximum height reached by the ball?
the maximum height reached by the ball is 2m
d) At what time(s) will the ball be 15 meters in the air?
H=+15m
15=+-5t%5E2%2B20t ....solve for t
15%2B5t%5E2-20t=0
5t%5E2-20t%2B15=0
5%28t%5E2-4t%2B3%29=0...will be zero if
t%5E2-4t%2B3=0....factor
t%5E2-3t-t%2B3=0
%28t%5E2-3t%29-%28t-3%29=0
t%28t-3%29-%28t-3%29=0
%28t+-+3%29+%28t+-+1%29+=+0
=> t=3 , t=1