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Sagot :
We can use the conservation of energy:
[tex]E1=E2[/tex]Where:
E1 = Energy of the car in the first hill
E2 = Energy of the car in the second hill
So:
[tex]\begin{gathered} K1+U1=K2+U2 \\ where: \\ K1=\frac{1}{2}mv1^2=\frac{1}{2}m(2.7)^2=3.645m \\ U1=mgh1=9.8(12)m=117.6m \\ K2=\frac{1}{2}mv2^2=\frac{1}{2}m(5.5)^2=15.125m \\ U2=mgh2=9.8mh2 \end{gathered}[/tex]Therefore:
[tex]3.645m+117.6m=15.125m+9.8mh2[/tex]Divide both sides by m:
[tex]\begin{gathered} 3.645+117.6=15.125+9.8h2 \\ 121.245=15.125+9.8h2 \\ 106.12=9.8h2 \\ h2=\frac{106.12}{9.8} \\ h2\approx10.829m \end{gathered}[/tex]Answer:
The height of the second hill is approximately 10.829m
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