1. ## word problem help

Hi, can someone help me with this? I don't even know how to set up the equation.

If an object is propelled upward from a height of 96 feet at an initial velocity of 80 feet per second, then its height h after t seconds is given by the equation h = - 16t2 + 80t + 96. After how many seconds does the object hit the ground?

2. You already have the necessary equation: $h(t)=-16t^2+80t+96$. Hitting the ground means the height at which the object is is $0$: $h(t_0)=0$ where $t_0$ is the time it hits the ground (assuming it gets launched at $t=0$).

3. Originally Posted by EandH
Hi, can someone help me with this? I don't even know how to set up the equation.

If an object is propelled upward from a height of 96 feet at an initial velocity of 80 feet per second, then its height h after t seconds is given by the equation h = - 16t2 + 80t + 96. After how many seconds does the object hit the ground?
$h=-16t^2+80t+96$

as this is the height equation,
then when the ball first hits the ground, the height will be zero.
If you write the equation as a multiplication (factorise), the answer will be in one of the factors.

$h=0\ \Rightarrow\ -4t^2+20t+24=0\ \Rightarrow\ t^2-5t-6=0$

$(t-6)(t+1)=0$

This is zero for t=6 seconds and t=-1 second.
The t=-1 comes from the graph which can show times "preceding" when the ball was propelled!!

Hence the answer is 6 seconds.