## Related questions with answers

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 $y=-16.1 t^{2}+4.75$ and $y=-0.0042 x^{2}+4.75$ 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.

Solution

VerifiedCargo 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 car got opropel forward. Use the equations,

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

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 $t$ we will use that is $y=0$ because the cargo will height $0$ after impact to the ground. So. we have,

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

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