Question

A baseball is thrown upwards from a height of 5 feet with an initial speed of 64 feet per second, and it's height h (in feet) from the ground is given by h(t)= 5 + 64t-16t^2 where t is time in seconds. Determine at what time the ball reaches its maximum height.

Answer 1

There are many ways to find this answer. Perhaps the simplest is to realize how acceleration relates to velocity.

a=(vf-vi)/(t2-t1)

Now for this problem we know that the ball reaches its maximum height when v=0, because this is the point when velocity is changing from positive to negative, hence it is at its highest position. And we also know that the acceleration is only cause by gravity or about -32 ft/s^2.

So in general we know that a=g=-32, vf=0, vi=64, t1=0 and t2 is the time when the ball is at its maximum height so:

a=(vf-vi)/(t2-t1) becomes

g=-vi/t2

t2=-vi/g

t=-vi/g, and in this case

t=-64/-32

t=2 seconds

In general for any quadratic of the form ax^2+bx+c=y, the x-coordinate of the vertex, which is the point of either an absolute minimum or maximum for the quadratic:

x=-b/(2a)

Which again we can see in this case is:

t=-64/(2*-16)

t=-64/-32

t=2 seconds