Question

37 multiple choice 2 points a ball is thrown off a 400 - foot - tall building at a velocity of 45 ft/sec. the function ( f(x)=-16 x^{2}+45 x + 400 ) (where ( x ) is the number of seconds elapsed) models the height of the ball after the toss. how high is the ball 2 seconds after it is thrown?

Explanation:

Step1: Substitute \(x = 2\) into the function

We have the function \(f(x)=-16x^{2}+45x + 400\). Substitute \(x = 2\) into it: \(f(2)=-16\times(2)^{2}+45\times2 + 400\).

Step2: Calculate each term

First, calculate \(-16\times(2)^{2}\): \(-16\times4=-64\).
Second, calculate \(45\times2 = 90\).

Step3: Sum up the terms

Now, \(f(2)=-64 + 90+400\).
\(f(2)=(-64 + 90)+400=26+400 = 426\).

Answer:

The ball is \(426\) feet high \(2\) seconds after it is thrown.