(a)The function 𝑓(𝑡)=−5𝑡2+20𝑡+60f\left(t\right)=-5t^2+20t+60f(t)=−5t 2 +20t+60 models the approximate height of an object 𝑡tt seconds after it is launched. How many seconds does it take the object to hit the ground? (Enter only the number)
Question
(a)The function 𝑓(𝑡)=−5𝑡2+20𝑡+60f\left(t\right)=-5t^2+20t+60f(t)=−5t 2 +20t+60 models the approximate height of an object 𝑡tt seconds after it is launched. How many seconds does it take the object to hit the ground? (Enter only the number)
Solution
The object hits the ground when the height is zero, i.e., when f(t) = 0.
So, we need to solve the equation -5t^2 + 20t + 60 = 0 for t.
This is a quadratic equation in the form of at^2 + bt + c = 0, where a = -5, b = 20, and c = 60.
We can solve this equation using the quadratic formula: t = [-b ± sqrt(b^2 - 4ac)] / (2a)
Substituting the values of a, b, and c into the formula, we get:
t = [-20 ± sqrt((20)^2 - 4*(-5)60)] / (2-5) t = [-20 ± sqrt(400 + 1200)] / -10 t = [-20 ± sqrt(1600)] / -10 t = [-20 ± 40] / -10
This gives us two solutions: t = 2 and t = -6.
However, time cannot be negative, so we discard the negative solution.
Therefore, it takes the object 2 seconds to hit the ground.
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