Respuesta :
Answer:
2 seconds
Step-by-step explanation:
You are given an equation, a height, and a velocity. Using this you need to solve for t:
[tex]384=-16t^2+160t\\16t^2-160t+384=0\\t^2-10t+24=0\\(t-6)(t-4)=0\\t=4, 6[/tex]
We get 2 answers for t. This is because the ball is at 384ft twice - once on its way up, and again on its way down. The ball is at (or above) 384ft for 6 - 4 = 2 seconds.
Answer:
4 and 6 seconds
Step-by-step explanation:
We have the following information about the problem:
initial velocity: [tex]v_{0}=160ft/s[/tex]
height: [tex]h=384ft[/tex]
And the projectile formula is:
[tex]h=-16t^2+v_{0}t[/tex]
substituting known values
[tex]384=-16ft^2+160t[/tex]
To simplify the equation we divide both sides by 16:
[tex]\frac{384}{16} =\frac{-16}{16}t^2+\frac{160}{16}t[/tex]
[tex]24=-t^2+10t[/tex]
Now, we move all terms to the left:
[tex]t^2-10t+24=0[/tex]
And we have a quadratic equation for the time that can be solved by factoring.
To factor we open two parenthesis and put [tex]t[/tex] o each, and we look for two numbers that multiplied result in [tex]+24[/tex] and added together result in [tex]-10[/tex]. Those numbers are [tex]-4[/tex] and [tex]-6[/tex] (because (-4)(-6)=24 and -4+(-6)=-10)
So the factorization is as follows:
[tex](t-4)(t-6)=0[/tex]
and by the zero product property (if two terms when multiplied result in zero, one of them or both are equal to zero):
[tex](t-4)=0[/tex]
⇒ [tex]t=4[/tex]
[tex](t-6)=0[/tex]
⇒ [tex]t=6[/tex]
The times that the ball is at a height of 384 ft are 4 and 6 seconds.
