Question

A particle is moving along a projectile path at an initial height of 160 feet with an initial speed of 144 feet per second. This can be represented by the function H(t) = −16t2 + 144t + 160. What is the maximum height of the particle?

Answer 1

484 feet

Explanation:

First, look at H(t) = −16t^2 + 144t + 160 as ax^2 + bx + c

Second, write out a=, b=, c=, which are a=-16, b=144, c=160

Third, plug a, b, & c into the equation for vertex which is t= -b/(2a) or t=-144/[2*(-16)], t=4.5.

Fourth, plug t=4.5 into the original equation H(t) = −16t^2 + 144t + 160, so H(4.5)=−16((4.5)^2) + 144(4.5) + 160 = -324 + 648 +160 = 484 feet

Answer 2

The maximum height of the particle is 484 m.

Explanation:

Given that,

A particle is moving along a projectile path at an initial height of 160 feet with an initial speed of 144 feet per second. This can be represented by the function :

`H(t) = -16t^2 + 144t + 160` ....(1)

We need to find the maximum height of the particle. For maximum height put
`(dH)/(dt)=0`

So,

`(d(-16t^2+144t+160))/(dt)=0\\\\-32t+144=0\\\\t=4.5\ s`

Put t = 4.5 s in equation (1) as :

`H(t) = -16(4.5)^2 + 144(4.5)+ 160\\\\H(t)=484\ m`

So, the maximum height of the particle is 484 m.