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The equation y = -4.2t^2 + 42t + 18.9 describes the height (in meters) of a ball tossed up in the air at 42 meters per second from a height of 18.9 meters from the ground, as a function of time in seconds. In how many seconds will the ball reach maximum height?

 Feb 1, 2022
 #1
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The maximum value is \(18.9 - {1764\over-16.8} = 123.9\). Subsitute the equation like this:\(123.9=-4.2t^2+42t+18.9\). Solve for \(t\), and you get \(\color{brown}\boxed 5\)

 Feb 1, 2022
 #2
avatar+118609 
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The equation y = -4.2t^2 + 42t + 18.9 describes the height (in meters) of a ball tossed up in the air at 42 meters per second from a height of 18.9 meters from the ground, as a function of time in seconds. In how many seconds will the ball reach maximum height?

 

This is a concave down parabola, the max will be when t = axis of symmetry

 

max height will be reached when t = -42/-8.4 = 5seconds

 

Just as BuilderBoi already said  cool

 Feb 1, 2022
edited by Melody  Feb 1, 2022
 #3
avatar+2666 
0

The x-value of the vertex can be also be found by doing \(-b\over2a\), in a quadratic equation with the format \(a^2 + bx + c = 0\).

 

This also yields \(\color{brown}\boxed5\)

BuilderBoi  Feb 1, 2022
edited by BuilderBoi  Feb 1, 2022

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