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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.

2 Answers

7 votes

Answer:

The arrow reach it maximum height after 2 seconds.

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=((-b)/(2a),f((-b)/(2a)))

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


(-b)/(2a)=(-(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.

User Jared Joke
by
8.9k points
3 votes

Answer:

After 2 seconds

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 ⇒
(32t)/(32)=
(64)/(32)

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

User Simone Sessa
by
8.2k points
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