Asked by Anonymous
a car is pared on a cliff overlooking the ocean on an incline that makes an angle of 24 below the horizontal. the negligent driver leaves the car in a neutral and emergency brakes are defective. the car rolls down the incline with a constant acceleration of 4 m/s^2 for a distance of 50 m to the edge of the cliff, which is 30 m above the ocean. find: the car's position relative to the base of the cliff when the car lands in the ocean and the length of time the car is in the air. physics
Answers
Answered by
Damon
well, first the roll
d = (1/2)a t^2
50 (1/2)(4) t^2
t^2 = 25
t = 5 seconds conveniently
speed = a t = 4*5 = 20 m/s at start of dive
so u = 20 cos 24 for the whole trip down
=18.3 m/s
Vi = -20 sin 24 = initial vertical speed = - 8.13 m/s
H = Hi + Vi t - 4.9 t^2
0 = 30 - 8.13 t -4.9 t^2
solve quadratic for t, time falling
in air then
distance from cliff = u t
d = (1/2)a t^2
50 (1/2)(4) t^2
t^2 = 25
t = 5 seconds conveniently
speed = a t = 4*5 = 20 m/s at start of dive
so u = 20 cos 24 for the whole trip down
=18.3 m/s
Vi = -20 sin 24 = initial vertical speed = - 8.13 m/s
H = Hi + Vi t - 4.9 t^2
0 = 30 - 8.13 t -4.9 t^2
solve quadratic for t, time falling
in air then
distance from cliff = u t