A projectile is fired upward from ground level with an initial velocity of 320 per second. (Assume t=0seconds corresponds to the time the object is fired. Use 32ft/sec^2 as acceleration due to gravity.) (s=-32t^2+vOt+sO) At what instant will it be back at ground level? When will the height exceed 1584 feet?

Question

xishem · Answer

Do you know what s represents in the equation?

anonymous · Answer

S= hieght of object

xishem · Answer

Alright, so ground level is 0 height. If you plug in 0 for the height and solve for time, you will find the instant at which the projectile is back at ground level.

Keep in mind that you'll get two answers since it is at ground level when you fired it (t=0) and at another time later on.

anonymous · Answer

0=32t^2+320+0?

xishem · Answer

Close, just a few mistakes. Be careful about signs and variables:$$0=-32t^2+320t+0$$

anonymous · Answer

so how do you figuire out the time its back at ground level?

xishem · Answer

You've put in the conditions, you just need to find the values of t that satisfy the equation.

anonymous · Answer

-32t(t+10)
t=-10
t=0

anonymous · Answer

sorry negative to t=10

anonymous · Answer

1584 <-32t^2+320t
0<-32t^2+320t-1584
Then I just use the quadratic formula to figuire it out right?

xishem · Answer

That's right.

Yep.

anonymous · Answer

Kay thank you. :) you are the best. <3

xishem · Answer

Are you having troubles?

anonymous · Answer

sorry it redid that

dan815 · Answer

umm hey

dan815 · Answer

you got a mistake in your equation

anonymous · Answer

my labtop is being weird but i got the answer $$(-\sqrt{10t+\frac{ 99 }{ 2 }}, \sqrt{10t +\frac{ 99 }{ 2 }})$$

dan815 · Answer

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