A ball is launched straight up from the ground with an initial velocity of 128 feet per second.

Question

a. At what time does the ball reach its maximum height?

 

b. Find the maximum height of the ball:

Answer 1

Use the equation for modeling the height of a projectile as a function of time. h(t) = -16t^2 + Vo(t) + ho. Where, Vo is the initial speed of the projectile and ho is the initial height. Here, Vo = 128ft/s and the initial height is zero. Plugging in those values yields: 

 

h(t) = -16t^2 + 128t 

 

since this equation is quadratic, it's maximum must occur at its VERTEX!! Therefore, t = -b/2a = -128/2(-16) =-128/-32= 4s.  The time to reach the max height is 4 seconds. If you were asked to calculate the max height you would need to plug in 4 into h, in other words evaluate h(4). 

 

The max or min of ANY quadratic function is equal to the y coordinate of its vertex. The x coordinate tells when or where the max or min occurs.

 

the vertex is an ordered pair defined by:

(-b/2a, f(-b/2a) 

Answer 2

v=u -gt

Final velocity at maximum height =0

Initial velocity =128 Ft/sec

g=32ft/sec^2

 

0=128-32t

t= 4 sec

 

Maximum height

h=ut-(1/2)gt^2

 

h=128*4-0.5*32*(4^2)

 

  =512-256=

   =256 ft