Question

A particle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second. What is the maximum height of the particle?

Hint: Use H(t) = -16t2 + vt + s.
128 feet
224 feet
272 feet
484 feet

Answer 1

D) 484 feet

Explanation:

Given:
`H(t) = -16t^2 + vt + s`, where t is the time, v is the initial velocity/speed and s is initial height.

and

v = 144 feet and s = 160 feet

Now plug in the given values in the given
`H(t) = -16t^2 + vt + s`

`H(t) = -16t^2 + 144t + 160`

This is in the form of quadratic function
`f(x) = ax^2 + bx + c, where a \\eq 0`

The maximum height attain in vertex. Here we have formula to find the vertex of the x coordinate.

x =
`(-b)/(2a)`

Here b = 144 and a = -16

So, t =
`(-144)/(2*-16) = (-144)/(-32) = 4.5`

The maximum height attain at t = 4.5 seconds. To find the maximum height plug in t = 4.5 in
`H(t) = -16t^2 + 144t + 160`

H(4.5) =
`-16(4.5)^2 + 144 (4.5) + 160\\`

= -16(20.25) + 648 + 160

= -324 + 648 + 160

= -324 + 808

= 484 feet

So the maximum height of the particle is 484 feet.