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a ball is thrown straight upward from a bridge and then falls all the w…

Question

a ball is thrown straight upward from a bridge and then falls all the way to the ground (past the bridge). the balls height ( h ) (in feet) above the ground ( t ) seconds after the ball is thrown is given by ( h(t)=-16 t^{2}+48 t + 113 ). what is the meaning of ( h(4) )?
after 4 seconds, the height of the ball is 241 feet above ground.
after 4 seconds, the height of the ball is 64 feet above ground.
after 4 seconds, the height of the ball is 3791 feet above ground
after 4 seconds, the height of the ball is 113 feet above ground.
after 4 seconds, the height of the ball is 49 feet above ground.

Explanation:

Step1: Substitute \(t = 4\) into the function

We have \(h(t)=-16t^{2}+48t + 113\). Substitute \(t = 4\) into it:

$$ LATEXBLOCK0 $$

Step2: Calculate \(-16\times(4)^{2}\)

First, calculate \((4)^{2}=16\), then \(-16\times16=-256\)

$$ h(4)=- 256+48\times4+113 $$

Step3: Calculate \(48\times4\)

\(48\times4 = 192\)

$$ h(4)=-256 + 192+113 $$

Step4: Calculate the sum

\(-256+192=-64\), then \(-64 + 113=49\)

$$ h(4)=49 $$

Answer:

After 4 seconds, the height of the ball is 49 feet above ground.