Question

A ball is thrown vertically upward. After seconds, its height (in feet) is given by the function 56t - 16t^2 . What is the maximum height that the ball will reach?

Answer 1

Answer:

Therefore the maximum height that the reach is 49 feet after 1.75 seconds.

Step-by-step explanation:

Given that,

A ball is thrown vertically upward. After t seconds its height is given by the function

$h=56t-16 t^2$

where h is height in feet.

Maximum value:

Given function f(x)= at²+bt+c.

1. Find the value $-\frac{b}{2a}$
2. Putting $x= -\frac{b}{2a}$ in the given function.
3. The maximum value of the given function is $f(-\frac{b}{2a})$.

Here a= -16 ,  b=56 and c=0

The ball attains its maximum height when $t=-\frac{56}{2.(-16)}=1.75$ s.

Putting the value of t in the given function

$h=(56\times 1.75)-16(1.75)^2$

  =49 feet

Therefore the maximum height that the reach is 49 feet after 1.75 seconds.