A car is parked on a cliff overlooking the ocean on an incline thatmakes an angl

Question

A car is parked on a cliff overlooking the ocean on an incline thatmakes an angle of 17.0° below thehorizontal. The negligent driver leaves the car in neutral, and theemergency brakes are defective. The car rolls from rest down theincline with a constant acceleration of 2.75 m/s2 for a distance of35.0 m to the edge of the cliff, whichis 30.0 m above the ocean. Find thefollowing. (a) The car's position relative to the base ofthe cliff when the car lands in the ocean.
m
(b) The length of time the car is in the air.
s (a) The car's position relative to the base ofthe cliff when the car lands in the ocean.
m
(b) The length of time the car is in the air.
s

Explanation / Answer

The first step is to find the velocity at the bottom of theincline. This is done using the following:       vf2 = vo2 + 2 a x   wherethe original speed at the top of the incline is given to be zerosince the car was at rest. so,    vf = [(2)(2.75m/s2)(35m)] =  13.87 m/s.   This is the initial velocityvo as the car leaves the incline and begins itsparabolic projectile trajectory. As in any projectile problem, the first steps are to find thex- and y-components of vo . vox = (13.87 m/s) cos 17 = 13.26m/s   and this component is constant voy = - (13.87) sin17   = - 4.055 m/s In the vert direction:     y =voy t - 1/2 g t2                              -30m = - 4.055 t - 4.9 t2 rearrange to get       4.9 t2 + 4.055 t - 30 = 0  This is a standard-looking quadratic eqn that can be solved by anyof the following three methods: (1) quadraticformula,   (2) graphing calculator; trace; or CALC: ZEROto find roots, or (3) some calculators will solve quadratics at the push of abutton. At any rate,   t = 2.095 sec. finalanswer The horizontal distance, x, from the edge of the cliff isgiven by: x = vox t   x = (13.26 m/s) (2.095 sec) =   27.78 m   final answer.

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