tastecraftedmcd.com->Customizable Word problem Solvers ->Travel-> SOLUTION: A vehicle is parked on a cliff overlooking the oceanon one incline that renders an angle of 18.0◦below the horizontal. The negligent driverleaves the car in neutral, and also the e var visible_logon_form_ = false;Log in or register.Username: Password: it is registered in one easy step!.Reset your password if friend forgot it."; return false; } "> log On Ad: over 600 tastecraftedmcd.com Word troubles at edhelper.com


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Click here to check out ALL difficulties on take trip Word ProblemsQuestion 214250: A automobile is parked on a cliff overlooking the oceanon an incline that makes an edge of 18.0◦below the horizontal. The negligent driverleaves the vehicle in neutral, and the emergencybrakes space defective. The auto rolls indigenous restdown the incline v a continuous accelerationof 2.00 m/s2 and travels 43.0 m to the sheet ofthe cliff. The cliff is 40.0 m above the ocean.The acceleration of gravity is 9.81 m/s2 .a) just how long is the automobile in the air? Answerin units of s.b) What is the car�s position relative come thebase the the cliff when the automobile lands in theocean? price in units of m.whats actions would u take it to settle this problem Answer by Alan3354(67465)
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(Show Source): You can put this systems on her website! A auto is parked top top a cliff overlooking the oceanon one incline that renders an angle of 18.0◦below the horizontal. The negligent driverleaves the automobile in neutral, and the emergencybrakes room defective. The vehicle rolls indigenous restdown the incline v a consistent accelerationof 2.00 m/s2 and also travels 43.0 m to the sheet ofthe cliff. The cliff is 40.0 m above the ocean.The acceleration of gravity is 9.81 m/s2 .a) exactly how long is the vehicle in the air? Answerin units of s. B) What is the car�s place relative come thebase that the cliff as soon as the vehicle lands in theocean? answer in systems of m. -----------------------------s = (at^2)/243 = (2*t^2)/2t = sqrt(43) secs to go the 43 metersv = at = 2sqrt(43) m/sec when the car leaves the incline and also is airborne.v =~ 13.114877 m/sec------------The vertical component of v is v*sin(18) = -4.05272 m/sec (neg because is down)The horizontal part of v is v*cos(18) = 12.473 m/sec---------------------h(t) = -9.81t^2 - 4.05272t + 40h(0) = 40 (when it leaves the ground)Find t in ~ h = 0 (when it access time the water)-9.81t^2 - 4.05272t + 40 = 0 solved by pluggable solver: solve quadratic equation (work shown, graph etc)Quadratic equation (in our instance ) has actually the complying with solutons: because that these services to exist, the discriminant must not it is in a an adverse number. First, we must compute the discriminant : . Discriminant d=1586.0245393984 is higher than zero.

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That way that there space two solutions: . Quadratic expression deserve to be factored: Again, the price is: -2.23637327194893, 1.8232519671579.Here"s your graph:t =~ 1.823252 seconds----------------------The horizontal component of v is v*cos(18) = 12.473 m/sec12.473 m/sec * 1.823252 secs =~ 22.741 meter from the base of the cliff
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