How can I calculate the distance a car will travel after driving off a cliff?

[antphany]

Homework Statement

A 500kg car drives off a cliff at 50km/h how far would it land from the base of the cliff if it was in the air for 10 seconds

Homework Equations

F=ma
F=mg
d=1/d at^2

The Attempt at a Solution

i first did f=mg to find f
then i did f=ma converted that to find a
then did 1/2 at^2 converted that 2
d x 0.5 / 9.8 then i root it to find t
but i got the wrong answer, the answer is 9 seconds can some one tell me what i did wrong

 

[Super Luminal]

I would think the the height of the cliff is a relevant variable, and that the 10 sec. airborne time of the car is at best irrelevant and at worst intentionally confusing. For sake of argument, if the car went off say, a 10,000 foot sheer drop it would spend more than ten seconds in the air...if the "cliff" was little more than an embankment then the car would probably spend less then 10 seconds in the air...both scenarios would render the question moot. The real question that should be asked, the way I see it is like this, "If a 500kg car goes of a cliff at 50 km and is airborne for ten seconds THEN how high was the cliff it went off?"

I am a layman, a noob, but physics enthusiast...but that's how I see it.

 

[Doc Al]

but i got the wrong answer, the answer is 9 seconds can some one tell me what i did wrong

How can 'the answer' be 9 seconds? You were told that it was in the air for 10 seconds.

What you are asked to find is how far from the cliff it landed.

 

[da_steve]

assuming no loss for friction

horizontal velocity = 50km/h

time = 10 secs

distance = vel*time

 

[jetwaterluffy]

i first did f=mg to find f
then i did f=ma converted that to find a

This step is a waste of time, unless you are going to factor in an air resistance force. g describes the accerleration, so there is no need to turn it into a force and back again.