A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24 degrees 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 2.4m/s^2 and travels 52m to the edge of the cliff. the cliff is 28m above the ocean.
How long is the car in the air? the acceleration of gravity is 9.81m/s^2. answer in units of s
What is the cars position relative to the base of the cliff when the car lands in the ocean? answer in units of m
How long is the car in the air? the acceleration of gravity is 9.81m/s^2. answer in units of s
What is the cars position relative to the base of the cliff when the car lands in the ocean? answer in units of m
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Velocity at which it launches = sqrt. (2ad), = 15.8m/sec.
Horizontal component = (cos 24) x 15.8, = 14.43m/sec.
Vertical component = (sin 24) x 15.8, = 6.426m/sec.
To gain this vertical velocity, the car could have dropped a distance of (v^2/2g) = 2.1 metres.
Vertical velocity of car at ocean = sqrt. (2gh), where h = (28 + 2.1m), = 24.3m/sec.
Change in velocity since launch = (24.3 - 6.426) = 17.874m/sec.
Time in air = (v/g), = 17.874/9.81, = 1.82 secs.
Distance from cliff = (1.82 x 14.43) = 26.26 metres.
Horizontal component = (cos 24) x 15.8, = 14.43m/sec.
Vertical component = (sin 24) x 15.8, = 6.426m/sec.
To gain this vertical velocity, the car could have dropped a distance of (v^2/2g) = 2.1 metres.
Vertical velocity of car at ocean = sqrt. (2gh), where h = (28 + 2.1m), = 24.3m/sec.
Change in velocity since launch = (24.3 - 6.426) = 17.874m/sec.
Time in air = (v/g), = 17.874/9.81, = 1.82 secs.
Distance from cliff = (1.82 x 14.43) = 26.26 metres.