Does anyone know how to do this problem? The height, s, in feet of a ball thrown straight up is given by the function s(t)=-16t^2+64t, where t is the time in seconds. Find the time it takes the ball to reach its maximum height. What is the maximum height?

Question

Does anyone know how to do this problem?  The height, s, in feet of a ball thrown straight up is given by the function s(t)=-16t^2+64t, where t is the time in seconds.  Find the time it takes the ball to reach its maximum height.  What is the maximum height?

anonymous · Answer

64 feet..There are two ways of doing this.. 1. Treat like a quadratic. Make -16t^2 + 64t + 0 = 0. The max will be when x=-b/2a. x=-64/(2*-32) = 2. When x = 2, y = 64ft, in this case x and y are t and s(t). The other way is with calculus. The max is reached when dy/dx (derivative) is 0. dy/dx of -16t^2 + 64t is -32t + 64. We set this equal to 0 and get t=2. Plug in and get 64. dy/dx is obtained by taking all terms and doing the following: take the exponent of the term, times the coefficient. That's your new coefficient. Then take the x term and make the exponent one less. Notice how the plane x becomes 1, because x is x to the 1st power. x to the zero power is 1.

anonymous · Answer

your vertex x=-b/2a    x=t=-64/2(-16)=64/32=2sec
max ht s=-16(2)^2+64(2)=64ft

anonymous · Answer

are you in calculus class tburn?

anonymous · Answer

No this is for intermediate algebra.  I have a final tomorrow.  This is one of the review problems.  And I had no idea how to figure it out. Thank you so much for your help!  You are awesom!

anonymous · Answer

ok god luck tburn for you exam