Solving Falling Objects with Formulas: Help Needed!

[Spdmnky]

Hey guys, just wanted to thank everyone for some help (mainly from other posts) that have saved me quite a headache in the past. Now I am having some major issues with falling objects and formulas and such... I missed about 4 days of class and am wayyyyyyy behind. So, with that said here we go...

A stone is thrown vertically upward with a speed of 13.5 m/s from the edge of a cliff 66.0 m high. What is the speed of the stone right before hitting the ground? What is the total distance the stone traveled? and lastly, How much later does it reach the bottom of the cliff?

I tried to use the formula:

Y_f=y_i +V_i(t)+1/2gt^2

I come up with what seems like random awnsers and it confuses me Any assistance would be greatly appreciated because this problem gives me one heck of a headache.

Thanks much,
Spd

 

[quasar987]

The trick is to separate the problems in 2 parts: 1) the ascension of the stone and 2) its falling.

You throw your stone upward vertically at 13.5 m/s. First let's define a convenient coorinate system. We will take the positive y axe as upward. During its trip upwards, there is a force $F=-mg$ acting on it. Newton's second law of motion says the force acting on an object equals its acceleration. So we have, $-mg = ma_y \Leftrightarrow a_y = -g = -9.80 \frac{m}{s^2}$. The acceleration of an object is the rate as which its speed changes. In mathematical form, $a_y=\frac{\Delta v_y}{\Delta t} \Leftrightarrow \Delta t = \frac{\Delta v_y}{-a_y} \ (*)$. And this is when you must use a bit of your physical intuition: at its maximum height, the speed of the stone is zero. So it starts at 13.5 m/s and end up at 0 m/s. This means that for the first part of the problem (as we have astuciously defined it), $\Delta v_y = v_y_f - v_y_i = 0 - 13.5 = -13.5 \frac{m}{s}$ We can use this result in equation $(*)$ to find the TIME that it took for our speed to drop from 13.5 m/s to 0. And now you have all the informations you need to use your equation $y_f=y_i+v_y_it + \frac{1}{2}gt^2$.

And for the second part of the problem, all of your final values from the first part become your initial value for the second part (because the second part is just the continuation IN TIME of the first part).

I think all you missed is the subtelty of dividing the problem in half, so you shouldn't have difficulty solving the second part.

 

[wais]

Hi Spd,

I'm glad to hear that previous posts have been helpful to you in solving problems. Falling objects and formulas can definitely be tricky, especially if you've missed a few days of class. Let's break down this problem step by step to help you understand it better.

First, let's identify what we know and what we're trying to find. We know that the stone is thrown vertically upward with a speed of 13.5 m/s and that the cliff is 66.0 m high. We're trying to find the speed of the stone right before hitting the ground, the total distance it traveled, and how much later it reaches the bottom of the cliff.

To solve this problem, we can use the formula you mentioned:

Y_f = y_i + V_i(t) + 1/2gt^2

where Y_f is the final position, y_i is the initial position, V_i is the initial velocity, g is the acceleration due to gravity (which is -9.8 m/s^2), and t is time.

Let's start by finding the time it takes for the stone to reach the ground. We know that the initial position, y_i, is 66.0 m and the final position, Y_f, is 0 m (since the ground is at 0 m). We also know that the initial velocity, V_i, is 13.5 m/s. Plugging these values into the formula, we get:

0 m = 66.0 m + (13.5 m/s)(t) + 1/2(-9.8 m/s^2)(t^2)

Simplifying this equation, we get:

0 = 66.0 + 13.5t - 4.9t^2

Now, we can use the quadratic formula to solve for t:

t = (-13.5 ± √(13.5^2 - 4(-4.9)(66.0)))/2(-4.9)

t = (-13.5 ± √(182.25 + 1293.6))/(-9.8)

t ≈ (-13.5 ± √1475.85)/(-9.8)

t ≈ (-13.5 ± 38.41)/(-9.8)

t ≈ (-52.91)/(-9.8) or (24.91)/(-9