Algebra
posted by Anonymous on .
A ball is thrown from an initial height of 2 meters with an initial upward velocity of 25m/s. The ball's height h (in meters) after t seconds is given by the following.
Find all values of t for which the ball's height is 12 meters.
Round your answer(s) to the nearest hundredth.

h(t) = 2+25t4.9t^2
h=12 when
2+25t4.9t^2 = 12
t= 0.44 and 4.66 seconds 
To find when the ball's height is
7
meters, we substitute
7
for
h
and solve for
t
.
=7+2−25t5t2
In order to solve for
t
, we first rewrite the equation in the form
=+at2+btc0
.
=+−5t225t50
Next, we use the quadratic formula to solve for
t
.
=t−b±−b24ac2a
Our equation has
=a5
,
=b−25
, and
=c5
.
Another way
We use these values in the formula.
t
=−−25±−−252·4·55·25
=25±52510
We get that
t
can be either of two values to solve the equation.
=t=−25525100.2087…
or
=t=+25525104.7912…
Rounding these values to the nearest hundredth, we get
=t0.21
or
=t4.79
.
So, the ball's height is
7
meters at approximately
0.21
seconds (on its way up) or
4.79
seconds (on its way down).