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
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
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