A ball is thrown upward with an initial velocity of 96 ft/sec from a height 640 ft. It's height h,in feet, after tseconds is given by h(t)= -16t^2 +96t+640. After how long will the ball reach the ground?
Tha ball will reach the ground in ___seconds

Question

A ball is thrown upward with an initial velocity of 96 ft/sec from a height 640 ft. It's height h,in feet, after tseconds is given by h(t)= -16t^2 +96t+640. After how long will the ball reach the ground?
Tha ball will reach the ground in ___seconds

anonymous · Answer

When it reaches the ground, $$\ h(t)=0$$

Therefore $$ -16t^2 +96t+640=0$$

anonymous · Answer

It is a quadratic equation, so there will be 2 solutions (which should be obvious to you, as there are 2 points when it touches the ground, before it was thrown and after it was thrown).

anonymous · Answer

With the question probably asking you for the later time.

anonymous · Answer

Oh...

anonymous · Answer

Do you understand this?

anonymous · Answer

Nope. All I got was h(t) = -16t^2 + 96t + 640 = 0 ==> t^2 - 6t - 40 = 0

anonymous · Answer

You understand why -16t^2 + 96t + 640 = 0, though? I'll help with the rest.

anonymous · Answer

Yeah

anonymous · Answer

t^2 - 6t - 40 is the same as (t-10)(t+4)

anonymous · Answer

So for (t-10)(t+4) to be =0, t =10 or =-4