Maximum height attained by arrow.

Question

An arrow is launched upward with a velocity of 96 feet per second from the top of a 60-foot stage. What is the maximum height attained by the arrow?

Answer 1

To find the maximum height attained by the arrow, we can use the kinematic equation:

vf^2 = vi^2 + 2ad

where vf is the final velocity (which is 0 at the maximum height), vi is the initial velocity (96 ft/s upward), a is the acceleration due to gravity (-32 ft/s^2 downward), and d is the displacement or change in height.

We want to find the displacement or change in height, which is the maximum height attained by the arrow.

Substituting the given values into the equation, we get:

0^2 = (96 ft/s)^2 + 2(-32 ft/s^2)d

Simplifying:

0 = 9216 ft^2/s^2 - 64d

64d = 9216 ft^2/s^2

d = 144 ft

Therefore, the maximum height attained by the arrow is 144 feet above the top of the 60-foot stage, or a total height of 204 feet above the ground.

Answer 2

h(t) = -16t^2 + 96t + 60

take the derivative and solve for t, then substitute back into the h(t) equation

h'(t) = -32t+ 96 = 0, t = 96/32 = 3 seconds

h(3) = -16(3)^2 + 96(3) + 60 = -144 + 288 + 60 = 204 feet maximum height

Answer 3

Firstly you need to write your equation.

So, assuming no air resistance,

-16t²+96t+60=0

this is a parabola. It is also a negative parabola so there with be a maximum point, the point is also the vertex. Vertex is (h,k)

x=h, h=-b/2a

So, -96/2(-16)=3, h=3 so x=3

we want to find y(or “k”)

to find this, we plug 3 into the equation for t

So -16(3)²+96(3)+60=204, k=204

the maximum height will be 204ft after 3 seconds