# Thread: Solve the equation. Round to the nearest tenth, if necessary.

1. ## Solve the equation. Round to the nearest tenth, if necessary.

If an object is projected upward with an initial velocity of 64 feet per second from a height of 336 feet, then its height t seconds after it is projected is defined by the expression h(t) = -16t2 + 64t + 336. How long after it is projected will it hit the ground?

THANK YOU!!!

2. Originally Posted by cechmanek32
If an object is projected upward with an initial velocity of 64 feet per second from a height of 336 feet, then its height t seconds after it is projected is defined by the expression h(t) = -16t2 + 64t + 336. How long after it is projected will it hit the ground?

THANK YOU!!!
Since h(t) is the height of the object we want h(t)=0 that is

$-16t^2+64t+336=0 \iff -16(t^2-4t-21)=0 \iff -16(t-7)(t+3)=0$

So t=7 or t=-3 since we are concerned with only future events we don't use t=-3. So the object hits the ground after 7 seconds.

Good luck.

3. Hello,

Originally Posted by cechmanek32
If an object is projected upward with an initial velocity of 64 feet per second from a height of 336 feet, then its height t seconds after it is projected is defined by the expression h(t) = -16t2 + 64t + 336. How long after it is projected will it hit the ground?

THANK YOU!!!
Find t such that $h(t)=0$.

$h(t)=-16t^2+64t+336=16(-t^2+4t+21)=-16(t-7)(t+3)$

$t=7$ or $t=-3$.

Because it's a nonsense to have a negative duration, the solution is 7seconds...

Edit :