Respuesta :
Answer:
The intervals of time will be [tex]\frac12 \ s[/tex] and 3 s.
Step-by-step explanation:
Given quadratic equation is
[tex]y=-16t^2+56t+2[/tex]
where y represents the height in feet, at a particular time, in second.
When a object is thrown upward direction,after a certain time it will attain the height 26 feet. Then when the object comes down it will again attain the height 26 feet.
To find the the time intervals, we put y= 26 in above quadratic equation.
[tex]\therefore26=-16t^2+56t+2[/tex]
[tex]\Rightarrow -16t^2+56t+2-26=0[/tex]
[tex]\Rightarrow -16t^2+56t-24=0[/tex]
[tex]\Rightarrow -8(2t^2-7t+3)=0[/tex]
[tex]\Rightarrow (2t^2-7t+3)=0[/tex]
[tex]\Rightarrow 2t^2-6t-t+3=0[/tex]
[tex]\Rightarrow 2t(t-3)-1(t-3)=0[/tex]
[tex]\Rightarrow( 2t-1)(t-3)=0[/tex]
[tex]\Rightarrow 2t-1=0 \ or,\ t-3=0[/tex]
[tex]\Rightarrow t=\frac12, 3[/tex]
The intervals of time will be [tex]\frac12 \ s[/tex] and 3 s.
The intervals of time will the projectile be above 26 feet is [tex]\frac{1}{2}s[/tex] and 3 s.
Calculation of the intervals of time:
Since the equation should be
[tex]y = -16t^2 + 56t + 2[/tex]
Here y shows the height in feet.
Now
Here y = 26
So,
[tex]26 = -16t^2 + 56t + 2\\\\-16t^2 + 56t +2 - 26 = 0\\\\-16t^2 + 56t - 24 = 0\\\\-8(2t^2 -7t + 3) = 0\\\\(2t^2 -7t + 3) = 0\\\\[/tex]
2t(t - 3) - 1 (t - 3) = 0
(2t - 1) (t - 3) = 0
So, The intervals of time will the projectile be above 26 feet is [tex]\frac{1}{2}s[/tex] and 3 s.
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