14)
Answer:
Explanation:
We would apply newton's formula of motion which is expressed as
h = ut + 1/2gt^2
where
h is the height
u is the initial velocity
g is the acceleration due to gravity in feet per second^2 = -32ft/s^2
From the information given,
h = 56t - 1/2 x 32 x t^2
h = 56t - 16t^2
The quadratic equation is expressed as
h = - 16t^2 + 56t
where
h represents the height of the rocket in feet
t represents the time in seconds
By the time the rocket returns to the ground, h = 0. We would substitute h = 0 into the equation and solve for t. We have
0 = - 16t^2 + 56t
The standard form of a quadratic equation is expressed as
y = ax^2 + bx + c
By comparing both equations,
a = - 16
b = 56
c = 0
The general formula for solving quadratic equations is expressed as
[tex]\begin{gathered} x\text{ = }\frac{-\text{ b}\pm\sqrt{b^2-4ac}}{2a} \\ x\text{ = }\frac{-\text{ 56 }\pm\sqrt{56^2-4(-\text{ 16 }\times0)}}{2\text{ }\times-\text{ 16}} \\ \frac{-\text{ 56 }\pm56}{-\text{ 32}} \\ x\text{ = }\frac{-\text{ 56 - 56 }}{-32}\text{ or x = }\frac{-\text{ 56 + 56}}{-\text{ 32}} \\ x\text{ = 3.5 or x = 0} \end{gathered}[/tex]Replacing x with t, we have
t = 3.5
It will return to the ground after 3.5 seconds