What is the maximum height the ball reached and when does the ball return to the ground?

Question

The formula h(t)=-16t+48t+160 represents the height of a ball, T seconds after it is launched. What is the maximum height the ball reached and also when does the ball return to the ground?

Answer 1

First, we need to fix the equation.  The first term should have t² in it.

 

h(t) = -16t² + 48t + 160

 

It reaches maximum height at the vertex of the height-time parabola, which is at

 

t = -48/[2(-16)] s = 1.5 s

 

h_max = h(1.5) = -16(1.5²) + 48(1.5) + 160 = 196 ft

 

It reaches the ground when h(t) = 0

 

-16t² + 48t + 160 = 0

 

t² - 3t - 10 = 0

 

(t - 5)(t + 2) = 0

 

t = 5 or -2

 

Since time starts at t=0, discard -2, so it reaches the ground 5s after launch.

 

Answer 2

The max occurs at the average of the zeros.

We need the zeros anyway to find where the

ball hits the ground (height is zer0)

 

-16t^2 + 48t + 160 = 0

 

16t^2 - 48t - 160 = 0

t^2 - 3t - 10 = 0 <--- divides everything by 10

 

( t   -   5)( t  +   2 ) = 0

 

t+2=0 results in negative time measures.

 

t-5 = 0 ---> t = 5

 

The ball hits the ground in 5 seconds.

 

The average of the zeros is (5 + -2)/2 = 3/2 = 1.5

The max height is -16(1.5)^2 + 48(1.5) + 160 =

                           -16(2.25) + 48(1.5) + 160 =

                            -36 + 72 + 160

                              36 + 160

                                96

 

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Calculus fans, read on.....

 

Maximizing, the first derivative is zero when

0 = h'(t) = -32t + 48

 

t = -48/-32 = -3/2 = -1.5