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 20.0° below the horizontal. The negligent driverleaves the car in neutral, and the emergency brakes are defective.The car rolls from rest down the incline with a constantacceleration of 3.21 m/s2for a distance of 30.0 m to the edgeof the cliff, which is 20.0 m abovethe ocean. (a) Find the car's position relative to thebase of the cliff when the car lands in the ocean.
1 m

(b) Find the length of time the car is in the air.
2 s (a) Find the car's position relative to thebase of the cliff when the car lands in the ocean.
1 m

(b) Find the length of time the car is in the air.
2 s

Explanation / Answer

it seems to me that you have to do (b) first. Use the acceleration rate and distance to calculate the car'svelocity V as it leaves the cliff. V = sqrt (2 a X) It will leave with the same downward cliff angle as the cliffslope. Get the horizontal and vertical components of the velocity as itleaves the cliff. Consider distance positive mesured downwards. Vxo = V cos 20.0 Vyo = V sin 20.0 (b) Write an equation for the time T it takes to fall to theocean. Vyo*T + (1/2) gT^2 = 30.0 m. (Solve for t) (a) Horizontal position = Vxo*T How fast is it going when it goes airborne? v = a t d = (1/2) a t^2 30 = .5 ( 3.21) t^2 t^2 = 18.6916 t = 4.32 s v = a t = 3.21 * 4.32 = 13.9 m/s call horizontal component u = 13.9 cos 20 = 13.04 m/s This isconstant until we hit the bleak ocean call vertical component Vo = 13.9 sin 20 = 4.75 m/s which is ourstarting speed down. Do the vertical problem first to find how long it takes to fall 50m with an initial speed down of 4.75 and an acceleration down of9.8 m/s^2 x = Xo + Vo t + (1/2) a t^2 so 20 = 0 + 4.75 t + 4.9 t^2 4.9 t^2 + 4.75 t - 20 = 0 t = [ -4.75 ± sqrt (4.75^2 +4*4.9*20)]/9.8 =[-4.75 ± 20.36 ]/9.8 = 1.59 seconds in air horizontal distance = u t = 13.04 * 1.59 = 20.73 meters

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