Question

Chris shoots an arrow up into the air. The height (in feet) of the arrow is given by the function h(t) = - 16t2 + 64t + 112 where t is the time in seconds. After how many seconds does the arrow reach it maximum height? Round to the nearest tenth of a second if necessary.

Answer 1

Answer:

After 2 seconds

Step-by-step explanation:

First step is find an equation of velocity:

h(t)=-16t2+64t+112. The velocity function is derivated from position function, so we should do that.

v(t)= h´(t)= -16*2*t+64 ⇒ v(t)= -32t+64.

Now we know the arrow reach the maximun height when this velocity is equal to zero when t>0.

0=-32t+64 ⇒ 32t=64, split each member for 32 ⇒ $\frac{32t}{32}$=$\frac{64}{32}$

t=2. That means after 2 sec. the arrow reach the maximum height

Answer 2

Answer:

The arrow reach it maximum height after 2 seconds.

Step-by-step explanation:

Chris shoots an arrow up into the air. The height (in feet) of the arrow is given by the function

$h(t)=-16t^2+64t+112$

where t is the time in seconds.

In the given quadratic function, the leading coefficient is negative. It means it is a downward parabola. Vertex of a downward parabola is the point of maxima.

If a parabola is defined by $f(x)=ax^2+bx+c$, then the vertex of the parabola is

$Vertex=(\dfrac{-b}{2a},f(\dfrac{-b}{2a}))$

In the given function $a=-16,b=24, c=112$.

$\dfrac{-b}{2a}=\dfrac{-(64)}{2(-16)}=2$

x-coordinate of vertex is 2. It means the arrow reach it maximum height after 2 seconds.

Substitute t=2 in the given function.

$h(2)=-16(2)^2+64(2)+112=176$

After 2 seconds the maximum height of the arrow is 176.