Question

A baseball is thrown upwards from a height of 5 feet with an initial speed of 64 feet per second, and its height h (in feet) from the ground is given by h(t) = 5 + 64t – 16t2 as shown in the graph where t is time in seconds. Using a graphing calculator, at what time does the ball reach its maximum height?Select one of the options below as your answer: A. 0 B. 1 C. 2 D. 3

Answer 1

Click 2nd then calc then max then set the left and right bound the click enter for TI 84
anyway, find the vertex

the x value of the vertex of an equaton in form
y=ax^2+bx+c
xvalue of vertex is -b/2a

y=-16t^2+64t+5
-b/2a=-64/(-16*2)=-64/-32=2
this x value is also the t value or time

question is at what time is the highest?

answer is C. 2 seconds

Answer 2

For this case we have the following equation:

$h(t) = 5 + 64t - 16t^2$

To find the time when it reaches its maximum height, we must derive the equation.

We have then:

$h'(t)= 64-32t$

Then, equaling to zero and clearing the time we have:

$64-32t = 0$

$t =\frac{64}{32}$

$t = 2$

Answer:

the ball reaches its maximum height at:

C. 2

Answer 3

For this case we have the following equation:

$h(t) = 5 + 64t - 16t^2$

To find the time when it reaches its maximum height, we must derive the equation.

We have then:

$h'(t)= 64-32t$

Then, equaling to zero and clearing the time we have:

$64-32t = 0$

$t =\frac{64}{32}$

$t = 2$

Answer:

the ball reaches its maximum height at:

C. 2