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Question

Cargo is tied with rope on the roof of a 4.5-foot-tall car. The car is traveling down a road at 42 mph and hits a concrete barrier. The rope snaps, allowing the cargo to propel forward. Use the equations y=16.1t2+4.75y=-16.1 t^{2}+4.75 and y=0.0042x2+4.75y=-0.0042 x^{2}+4.75 where y represents height in feet, x represents horizontal distance in feet, and t represents time in seconds, to find the time it takes for the cargo to hit the ground and the horizontal distance it travels.

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Answered 2 years ago
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Cargo is tied with rope on the roof of a 4.54.5-foot-tall car. The car is traveling down a road at 4242 mph and hits a concrete barrier. The rope snaps, allowing the car got opropel forward. Use the equations,

y=16.1t2+4.75y=0.042x2+4.75\begin{align*} y=-16.1t^2+4.75\\ y=-0.042x^2+4.75\\ \end{align*}

where y represents height in feet,x represents horizontal distance in feet, and t represents time in seconds, to find the time it takes for the cargo to hit the ground and the horizontal distance it travels.To calculate tt we will use that is y=0y=0 because the cargo will height 00 after impact to the ground. So. we have,

0=16.1t2+4.754.75=16.1t2t2=4.7516.1=0.30t=0.30=0.55seconds\begin{align*} 0=-16.1t^2+4.75\\ -4.75=-16.1t^2\\ t^2=\dfrac{-4.75}{-16.1}=0.30\\ t=\sqrt{0.30}=0.55\,\,\text{seconds} \end{align*}

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