How many seconds for the ball to reach its maximum height? What is the maximum height?

Question

A ball is thrown off a 64ft platform with an initial velocity of 96 ft/sec. The height of the ball after t seconds is given by the function h(t)= 16t^2 + 96t +64. How many seconds for the ball to reach its maximum height? What is the maximum height?

Answer 1

First, I am fairly certain you meant h(t) = -16 t^2 + 96t + 64. Otherwise, there is no maximum height that this ball reaches! 

 

Try graphing this correct height function, wolframalpha makes this simple: 

 

http://www.wolframalpha.com/input/?i=plot+-16t%5E2+%2B+96t+%2B64+from+t%3D0+to+t%3D7

 

Notice this is a parabola! A parabola has the form

 

ax^2 + bx + c

 

The highest/lowest point on a parabola (depending on whether it bends upwards or downwards) happens at the x value x= -b/(2a)

 

Here, we have b=96 and a= -16, so we have

 

x = -b/(2a) = -96/ (-2* -16)= -96/(-32) = 3

 

So the highest point on the parabola is when x=3, or in our case t=3. So the ball reaches its maximum height after 3 seconds. What is that height? To figure that out, we need to plus this values of t into our function h(t); after all, it tells us the height at time t!

 

h(3)= -16(3^2) + 96* 3 + 64= -16(9)+ 288 + 64 = -144 + 288 + 64 = 208

 

So the maximum height of the ball is 208 ft. That would be quite the arm to throw that!