Answer:
Therefore,
3.5 s long was the rocket in the air.
Step-by-step explanation:
Given:
A rocket is launched from the ground.
Height function for Rocket is given as
[tex]y=-16x^{2}+56x[/tex]
Where,
y = height in function with x as time in second
To Find:
How long is the rocket in the air, x =?
Solution:
Expression is given
[tex]y=-16x^{2}+56x[/tex]
So for time in air put y = 0 in above Expression
[tex]0=-16x^{2}+56x[/tex]
[tex]16x^{2}-56x=0\\16x(x-3.5)=0\\16x=0\ or\ x-3.5 = 0\\x=0\ or\ x= 3.5[/tex]
x cannot be negative
∴ x = 3.5 sec.
Therefore,
3.5 s long was the rocket in the air.
