Question

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?
426 ft
374 ft
522 ft
554 ft

Explanation:

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

Given \(f(x)=-16x^{2}+45x + 400\), when \(x = 2\), we have \(f(2)=-16\times(2)^{2}+45\times2 + 400\).
First, calculate \(-16\times(2)^{2}\):
\(-16\times4=-64\).
Then, calculate \(45\times2 = 90\).

Step2: Calculate the value of \(f(2)\)

Now, substitute the values back into \(f(2)\):
\(f(2)=-64 + 90+400\).
\(f(2)=(-64 + 90)+400\).
\(26+400 = 426\).

Answer:

426 ft.