Hey everyone. I need some major help. Here is my question.
The height of a ball thrown up at 320 feet per second is given by:
h(t)=-16tsquared +320t+5. Find the maximum height the ball reaches. Please help!
Thanks
The ball goes up, reaches its maximum height, stops, then starts coming down again.
The operative word there is "stops".
So if you can find the value of h when t is equal to zero you're made.
From the equations of motion you've got
h = u t + 1/2 a t^2
where u = 320, a = -32 and h is the value you're trying to find. This matches the equation you've been given, except for the 5. So write it:
h - 5 = -16 t^2 + 320 t
The 5 signifies the extra 5 feet above the ground you start throwing the ball upwards from.
So now you can look at the equation of motion that goes:
v^2 = u^2 + 2 a (h-5)
You've got v, it's equal to zero (see above), and h is what you want to find.
So you have h-5 = (v^2 - u^2) / (2 a)
and plugging in the numbers you get
h-5 = (0 - 320^2) / (2 x (-32))
that should be enough, you should be able to work it out from there.