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A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 23.1 ◦ below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 2.52 m/s 2 and travels 37.5 m to the edge of the cliff. The cliff is 21.1 m above the ocean. Find the car’s position relative to the base of the cliff when the car lands in the ocean. The acceleration due to gravity is 9.8 m/s 2 . Answer in units of m.

Respuesta :

Answer:20.17 m

Explanation:

Given

Angle of inclination [tex]=23.1 ^{\circ}[/tex]

acceleration of  car[tex]=2.52 m/s^2[/tex]

Distance travel 37.5 m to the edge of the cliff

Velocity after travelling 37.5 m

[tex]v^2-u^2=2as[/tex]

[tex]v^2-0=2\times 2.52\times 37.5[/tex]

v=13.74 m/s

Now the car is launched at an angle of [tex]23.1^{\circ}[/tex] with the horizontal

Vertical distance traveled

[tex]21.1=vsin23.1t+\frac{gt^2}{2}[/tex]

[tex]4.905t^2+5.39t-21.1=0[/tex]

t=1.596 s

Thus car position relative to the base of the cliff[tex]=vcos23.1\times t[/tex]

[tex]=13.74\times cos23.1\times 1.596=20.17 m[/tex]

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