A batter hits a baseball upward with an initial speed of 96 feet per second. After how many seconds does the ball hit the ground? Use the formula h = rt − 16t2 where h represents height in feet and r represents the rate in feet per second.

Question

anonymous · Answer

in this example $r=96$ although usually it is called $v_0$

anonymous · Answer

i guess the batter hits the ball that is lying on the ground, because there is supposed to be an initial  height involved as well
no matter 
your job is to solve 
$$96t-16t^2=0$$ so you know how to solve it?

anonymous · Answer

"no" is an ok answer, if you do not now how i can walk you through it

anonymous · Answer

no

anonymous · Answer

ok fine

anonymous · Answer

thank you

anonymous · Answer

$$96t-16t^2=0$$ is what we have to solve
it is probably easier to multiply both sides by $-1$ i.e. change each sign and solve 
$$16t^2-96t=0$$

anonymous · Answer

$16t^2$ and $96t$ both have a common factor of $t$
they also both have a common factor of $16$ because $16\times 6=96$

anonymous · Answer

so you can factor this as 
$$16t(t-6)=0$$

anonymous · Answer

so t=6?

anonymous · Answer

one answer is $16t=0\iff t=0$ but that is not your answer, that just says at the beginning the ball is on the ground
the other is 
right

anonymous · Answer

$t=6$ is the answer

anonymous · Answer

ty!

anonymous · Answer

yw