How to Solve Basic Kinematics Problems: Tips and Tricks

[don123]

Hey Guys, here are my questions (with the work included.) Please tell me what I am doing wrong as it is not working on webassign.

A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of a cliff 80.0 m high.

(a) How much later does it reach the bottom of the cliff?
s
(b) What is its speed just before hitting?
m/s
(c) What total distance did it travel?
m


Okay the answer to (c) is 102.9 m (it is the correct answer.)

For A (please open the attached picture if it helps.)

Basically I found the time from when the stone is thrown up to the time when the max height is reached. (velocity = 0) WIth that time, I doubled that to find the time until the 80 m cliff is reached. Then, I found the time it takes for the rock to reach the bottom of the cliff with the equation X= Vi(t) + .5at^2 (a = 9.8; x = 80; Vi = 15 and t = ?)

But unfortunately when I add up the two time measures, they are not working.

 

[don123]

I need help with another problem (also kinematics) and I do not want to double-thread, hence I am just posting it here.

A stone is dropped from the roof of a high building. A second stone is dropped 3.00 s later. How far apart are the stones when the second one has reached a speed of 14.0 m/s?

Basically, I found Vi, Vf, X, A, T for the first stone and for the second stone. For the second stone, I found out at what time it is that the 14m/s velocity is reached. From there, I used basic kinematics formulas and then subtracted the answers, but that also is wrong. Any help will be greatly appreciated.

 

[siddharth]

Your approach to solving the first problem should work. Have you done the math right? Have you used 'g' as the given in the question? (some times, the question may ask you to approximate 'g' as 10m/s^2)

For the second question, you could simply use the concept of relative velocities.

 

[Hootenanny]

Can you show your working for the additional question. It saves us working through the whole problem if the error is arithmerical.

 

[Doc Al]

Basically I found the time from when the stone is thrown up to the time when the max height is reached. (velocity = 0) WIth that time, I doubled that to find the time until the 80 m cliff is reached. Then, I found the time it takes for the rock to reach the bottom of the cliff with the equation X= Vi(t) + .5at^2 (a = 9.8; x = 80; Vi = 15 and t = ?)

As siddharth noted, this approach should work just fine. Show the details of what you did and we can check for errors. You can also solve this in a single step using:
$$y = y_0 + v_0 t + .5 a t^2$$
with the appropriate values for y, y_0, v_0, and a. (Mind the signs.)

For your second problem:

A stone is dropped from the roof of a high building. A second stone is dropped 3.00 s later. How far apart are the stones when the second one has reached a speed of 14.0 m/s?

Basically, I found Vi, Vf, X, A, T for the first stone and for the second stone. For the second stone, I found out at what time it is that the 14m/s velocity is reached. From there, I used basic kinematics formulas and then subtracted the answers, but that also is wrong. Any help will be greatly appreciated.

Again, nothing obviously wrong here, but the devil is in the details. Show how you found the time for the second stone to reach the stated speed, etc., and we can try and spot your error.