IT leaves with respect to the ground. That means it is going 30.8m/s with respect to the rail car.
distance=30.8*time
where time is the time to fall 1.5 m
An automobile with a mass of 1316 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 40,810 kg and is moving to the right at a constant speed v = 8.8 m/s on a frictionless rail. The automobile then accelerates to the left, leaving the railcar at a speed of 22 m/s with respect to the ground. When the automobile lands, what is the distance D between it and the left end of the railcar? See the figure.
3 answers
I tried to solve for the time using:
1.5=-1/2*(9.81)t^2
but I'm not getting the right answer.
1.5=-1/2*(9.81)t^2
but I'm not getting the right answer.
t=(1.5*2/g)^1/2