Answer:
-1) -4 (4t + 1) (t - 2) = 0
2) t = 2 sec
Step-by-step explanation:
The given equation is:
[tex]-16t^{2}+28t+8=0[/tex]
We need to factorize this equation.
Taking -4 as common from left hand side, we get:
[tex]-4(4t^{2}-7t-2)=0[/tex]
The term inside the brackets can be further simplified by breaking down -7t as -8t + t:
[tex]-4(4t^{2}-8t+t-2)=0\\\\ -4(4t(t-2)+1(t-2))=0\\\\ -4(4t+1)(t-2)=0[/tex]
This is the fully factored form of the given equation.
Using the zero product property,we can write:
4t + 1 = 0 → t = [tex]-\frac{1}{4}[/tex] → t = -0.25
t - 2 = 0 → t = 2
This means, the rocket touches the ground at time = -0.25 sec and at time = 2 sec. Since, negative time is not possible we neglect this value.
Therefore, we can conclude that the rocket launched by the child reaches/touches the ground after 2 seconds.
