Car Cliff Equation: Distance to Ground from Base

[billybobay]

Homework Statement

A car drives off a 50 m high cliff with a horizontal velocity of 20 m/s. How far from the base of the cliff will the car strike the ground? Assume there is no air resistance and the ground is level at the base of the cliff.


20 m

50 m

64 m

100 m

Homework Equations

20 m/s = square root -9.8/(2)(-50 m)
64.5 m
20 = (.31)(64.5)
20 = 20

The Attempt at a Solution

the car will strike the ground 64.5 or 64 meters away from the base of the cliff.

would that be correct and did I use the proper formulas?

 

[Delphi51]

I agree with your final answer to 2 digit accuracy.
I have no idea what formulas you used. You never wrote any.
Certainly "20 m/s = square root -9.8/(2)(-50 m)" is not a correct statement. The right side evaluates to 15.7, not 20.

 

[billybobay]

I agree with your final answer to 2 digit accuracy.
I have no idea what formulas you used. You never wrote any.
Certainly "20 m/s = square root -9.8/(2)(-50 m)" is not a correct statement. The right side evaluates to 15.7, not 20.

I used one of many equations which were given in my book. And changed my numbers from theirs to mine. They were using the equation to find the velocity so I just filled my velocity given to me in my question in and found for the change of x which in their equation they filled in. You would agree that it would be 64 meters though? What equation would you suggest would be a clearer one to use?

 

[Delphi51]

You are supposed to tell me what equations you used first! Always write an equation first, then fill in the numbers so your reader can tell what principle you are using and follow your work!

For horizontal and vertical motion problems, I always make two headings for horizontal and vertical. Under horizontal, there is no acceleration so I write the d = vt from grade 10 physics. Under vertical, there is the constant acceleration of gravity so I write the constant acceleration formulas V = Vi + at and d = Vi*t + ½at² from grade 11. Then I fill in every letter I can with the given numbers. If I find one of the 3 equations has only one unknown, I solve for it. Then use that value in the other equations as needed. Once in a while none of the three has only one unknown, so you have to solve two of the equations as a system of 2 equations with 2 unknowns. You can solve any trajectory problem this way.

Sorry about lecturing; it is late at night here. I taught this stuff for 30 years so it just falls out like a lecture. I will be impressed if you can write out a complete solution that a grade 11 student can follow. Then you can be a physics teacher, too.