Description
An object is launched from a platform. Its height in meters, yyy, is dependent on time in seconds after launch, xxx, and can be modeled with the function y=−5(x+1)(x−9) How many seconds after launch will the object hit the ground? ___seconds What is the maximum height that the object will reach? ___meters How many seconds after being launched will the object reach its maximum height? ___seconds What is the height of the object at the time of launch? ___meters
Explanation & Answer
1-To know how many seconds after launch the object will hit the ground, we have to solve the equation
y(x)=0 so x=-1 or x=9. The only possible answer is x=9 seconds as we remove the negative solution because x (time) is positive.
answer : 9 seconds.
2 - You have to find the maximum of the function y(x)=-5(x+1)(x-9)=-5(x^2-8x-9)
y'(x)=-5(2x-8). The maximum value is obtained for x as y'(x)=0 so x=4. to find the maximum height, we just have to calculate y(4)=125
answer : 125 meters
3 - Previously we calculated the value for x where y(x) is maximum. x=4. This is the number of seconds after beeing lauched where the object reach its maximum height.
answer : 4 seconds
4- To know the height of the object at the time of launch, we just have to calculate y(0). y(0) is the height of the object at x=0 or in other words at time of launch. y(0)=45
answer : 45 meters.
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