If you shoot an arrow vertically into the air from 5 ft with an initial velocity of 96 ft/second.?

Question

The height, h, in ft above the ground, at any time, t (in seconds) is modeled by h(t)=5+96t-16t^2

Deterimine the maximum height the arrow will attain.

The height, h, in ft above the ground, at any time, t (in seconds) is modeled by h(t)=5+96t-16t^2

A.) Deterimine the maximum height the arrow will attain.

B.) When will the arrow hit the ground?

Answer 1

The maximum height is given by the point where the slope (derivative) is zero.

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

So the maximum point will be at time t = 3.

h(3) = 5 + 96(3) - 16(3)²
h(3) = 5 + 288 - 16*9
h(3) = 293 - 144
h(3) = 149 ft.

Now just don't stand directly underneath this when the arrow returns to Earth!

Answer 2

The maximum value of a quadratic function at^2 + bt + c happens when t = -b/(2a)

Here t = -96/(2(-16))

Find this and plug it in for t to get the maximum height

Answer 3

Do your own math homework.