triggerhorse Posts: 8, Reputation: 1 New Member #1 Mar 22, 2009, 05:37 PM
Phy 11- Starts at rest at cliff edge If accel how far from bottom of cliff will land
Hi! A car starts at rest a certain distance (150m) away from edge of a cliff (64m high.) If car can accelerate at a certain rate (3.10m/s/s), how far from bottom of cliff will the car land?

I have know idea where to start on this question. Any help would be great.
Thanks
 sarnian Posts: 462, Reputation: 9 - #2 Mar 22, 2009, 06:10 PM
Hello triggerhorse

First you calculate how fast the car drives at the moment it leaves the cliff.
You have than a horizontal force (speed), a vertical force (gravity), and a vertical distance (64 meter). Calculate how long it takes to fall 64 meter.
Than calculate how far the car will travel forwards during that time.
Consider the air resistance nill.
Is that enough for you to calculate the distance?

Note : there is a Home --> Education --> Homework Help --> Math Board for this type of questions!
 triggerhorse Posts: 8, Reputation: 1 New Member #3 Mar 22, 2009, 06:31 PM
Originally Posted by sarnian
Hello triggerhorse

First you calculate how fast the car drives at the moment it leaves the cliff.
You have now a horizontal force (speed), a vertical force (gravity), and a vertical distance (64 meter). Calculate how long it takes to fall 64 meter.
Than calculate how far the car will travel forwards during that time.
Consider the air resistance nill.
Is that enough for you to calculate the distance?

Note : there is a Home --> Education --> Homework Help --> Math Board for this type of questions!
Hi! Thanks for the reply. How would I calculate "how fast the car drives at the moment it leaves the cliff?" If I know this, then I think I can work out the question from there.
 Perito Posts: 3,139, Reputation: 150 Ultra Member #4 Mar 22, 2009, 08:11 PM

The distance to the edge of the cliff is 150m. The car is traveling horizontally so there is no vertical component. It accelerates (initial velocity = 0)

You should know the formula,

$X = V_ot + \frac{1}{2}At^2$

where t is time

That's all you need to know.
 triggerhorse Posts: 8, Reputation: 1 New Member #5 Mar 22, 2009, 08:50 PM

Thank you
 sarnian Posts: 462, Reputation: 9 - #6 Mar 23, 2009, 05:55 AM
Hello triggerhorse

With all this help and assuming this allowed you to solve the question, please let us have your calculation how far away the car will land from the bottom of cliff.
 triggerhorse Posts: 8, Reputation: 1 New Member #7 Mar 23, 2009, 08:08 AM

final step for me was

d=vit+0.5a(tsquared)
d=30.4(3.6)+0
d=109.4m
Not sure if it is correct. Sounds logical.

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