Please answer the questions 001,002,003 Thanks A car is parked near a cliff over

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Please answer the questions 001,002,003

Thanks

A car is parked near a cliff overlooking the ocean on an incline that makes an angle of 21.7 degree with the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline and has a velocity 5 m/s when it reaches the edge of the cliff. The cliff is 19.4 m above the ocean. How far is the car from the base of the cliff when the car hits the ocean? The acceleration of gravity is 9.8 m/s^2 Answer in units of m/s. A particle at rest undergoes an acceleration of 2.9 m/s^2 to the right and 4.3 m/s^2 up. What is its speed after 6.9 s? Answer in units of m/s. What is its direction with respect to the horizontal at thus time? Answer between -180 degree and + 180 degree. Answer in units of degree. A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 9 m, y = 8.5 m, and has velocity upsilon_o = (9 m/s) i + (-7 m/s) j. The acceleration is given by a = (8.5 m/s^2) i + (6.5 m/s^2) j. What is the x component of velocity after 1s? Answer in units of m/s. What is the y component of velocity aft er 1 s? Answer in units of m/s. What is the magnitude of the displacement from the origin (x = 0 m, y = 0 m) after 1 s? Answer in units of m.

Explanation / Answer

given

theta = 21.7 degrees

vo = 5 m/s

h = 19.4 m

vox = 5*cos(21.7) = 4.64 m/s
voy = -5*sin(21.7) = -1.85 m/s

1) let t is the time taken for car to hit the water

use,

-h = voy*t - 0.5*g*t^2

-19.4 = -1.85*t - 0.5*9.8*t^2

on solving the above equation

t = 1.81 s

distance in horizontal direction before hitting the water,

x = vox*t

= 4.64*1.81

= 8.40 m

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