Quadratic equations answer asap please

Question

A toy rocket is launched
If a toy rocket is launched vertically upward from ground level with an initial velocity of 128 feet per second, then it's height h after t seconds is given by the equation h(t) =-16t2+128t(if air resistance is neglected)
a) how long will it take for the rocket to return to the ground?
b)after how many seconds will the rocket be 112 feet above the ground?
c)how long will it take the rocket to hit max height?
d)what is max height?

Answer 1

a) set h(t) = 0 and solve for t

 

-16t^2 + 128t = 0 implies 16t^2 = 128t. Dividing both sides by 16t yields the equation t = 8

 

It will take 8 seconds after launch for the rocket to return to the ground.

 

b) set h(t) = 112 and solve for t. 16t^2 -128t + 112 = 0. Divide both sides by 16 to obtain t^2 -8t + 7 = 0.

This can be factored into (t-7)(t-1)=0. Therefore, the rocket is 112 feet above the ground at t=1 and also t=7.

 

There are two times when the rocket is 112 feet above the ground, once after 1 second (on the way up), and once after 7 seconds (on the way down.)

 

c) To find the maximum height, set the first derivative equal to zero.  The first derivative of the equation 16t^2 - 128t + 112 = 0 is

32t -128 = 0 which implies t = 4

 

The rocket will reach its maximum height 4 seconds after launch.

 

d) simply use the answer from part c0 above in the original equation.  h(t) = -16t^2 + 128t

Now, h(t) = -16(4)^2 + 128(4) = -16*16 + 512 = -256 + 512 = 256

 

The maximum height is 256 feet.