suppose the function h(t)=-16t^2+50t+15 gives the height of an object in feet over time (in seconds).

Question

A. when does the object reach its highest point?

 

B. when does the object hit the ground?

Answer 1

Another, non-calculus, way to think of this problem is that the function in question describes a parabola.  So the object reaches the highest point in its trajectory at the parabola's maximum.

 

1. The parabola's function is described by: -16t² +50t +15; now we need to find this parabola's maximum.  For any parabola, the x-coordinate (or in this case, the t-coordinate) of the maximum (or minimum) is given by:

 

t = -b/2a where b = 50 and a = -16 for this function.

 

So, t = -50/-32 = 1.5625; the object is at its maximum height in 1.5625 seconds.

 

The previous answer does a good job of answer part B.

 

 

Answer 2

A. When you see "highest" or "maximum" think: "take the derivative and set it equal to zero." 

 

h'(t)=-32t+50

0=-32t+50

32t=50

t=50/32=25/16 seconds

 

You can check that this is a maximum and not a minimum by plugging t into h, and then trying values of t before and after 50/32 seconds and comparing the values for h (I highly recommend this check!)

 

B. The ground is height zero. So, set h(t)=0 and solve for t.

 

0=-16t^2+50t+15

0=16t^2-50t-15

Quadratic equation yields t = 3.04816 seconds

 

 

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