+0

# 1. An 2 kg object has an inital velocity of 4 m/s. If it comes to a stop in a distance of 12 meters.

+5
138
8

1. An 2 kg object has an inital velocity of 4 m/s. If it comes to a stop in a distance of 12 meters.

What was the acceleration of the object.

time is 12 / 4 = 3

12 / 3 = 4

would 4 be the acceleration?

2. If you throw a ball at 30 degrees into the air with a velocity of 18 m/s. How long is it in the air?

3. A 4kg object is 30m above the ground.

If it falls what is the object velocity 2 meters above the ground?

Guest Feb 26, 2017
Sort:

#1
+5541
0

For the first one, I'm not really sure but I think that this is how to do it.

Imagine a graph of the velocity of the object.

Its velocity is 4 m/s at x = 0, and its velocity is 0 m/s at x=12 meters.

So if y is the velocity and x is the distance, the equation of that line is: y = (-4/12)x + 4, or

y = (-1/3)x + 4.

The derivative of velocity is acceleration, so the acceleration is -(1/3) m/s/s.

You know the acceleration is going to be negative because the object is slowing down.

I'm not sure if this is right though because I didn't use the information about it being 2 kg at all.

hectictar  Feb 26, 2017
#2
+5541
+1

I also want to attempt the second one because why not haha I have no clue if I'm doing these right though.

So I'm going to work towards getting a position function of the ball.

Acceleration:

y''= -9.8 m/s2

Velocity:

y'=$$\int -9.8 dx$$

y' = -9.8x + c

18sin30° = -9.8(0) + c

c=9

y'= -9.8x + 9

Position:

y=$$\int(-9.8x+9)dx$$

y=$$\int-9.8x dx +\int9dx$$

y=$$-9.8\frac{x^{2}}{2}+9x$$

y= -4.9x2 + 9x

So that is the position function of the ball. So we need to find the two places where the ball touches the ground, aka where the y value equals 0.

0 = -4.9x2 + 9x

0 = x(-4.9x+9)

Set each factor equal to zero.

-4.9x+9 = 0 and x = 0

-4.9x=-9

x = 90/49

So y is 0 when x is 0 and when x is 90/49. The difference in those x values will be the distance it  traveled. So it traveled 90/49 meters. I just realized that we want the time it is in the air, not the distance traveled. Well it's okay the time it was in the air will just be 18 m/s divided by 90/49 meters. Which is 49/5 or 9.8 seconds. That was probably the longest possible approach to this problem but I kind of think I got the right answer.

hectictar  Feb 26, 2017
#5
+5541
0

Oh okay mine is wrong because I found the displacement not the distance, so when I tried to convert that to air time it just gave me something that was meaningless. I didn't really know what I was doing, heh.

But I'm confused though because when I tried to find the arc length from 0 to 90/49 I got about 8.61142 and then 18 divided by that is about 2.09. Howcome 2.09 seconds isn't the answer?

hectictar  Feb 26, 2017
#3
+79654
0

2. If you throw a ball at 30 degrees into the air with a velocity of 18 m/s. How long is it in the air?

We are using :

t  = 2 v0 sin θ / g

t =  2* 18m/s * sin (30)  / [9/8m /s^2]  =

36 * (1/2)  / 9.8    ≈  1.837 s

Here's a graph : https://www.desmos.com/calculator/2i436tqxnr

CPhill  Feb 26, 2017
#4
+79654
0

3. A 4kg object is 30m above the ground.

If it falls what is the object velocity 2 meters above the ground?

The time it takes to fall 28m  is :

-28  = -(1/2)9.81 * t^2

56/9.8  = t^2

sqrt (56/9.8)  = t ≈  2.39 s

And the velocity  is given by :

v(2.39)  =  9.8m/s^2 (2.39 s)   ≈ 23.4 m/s

CPhill  Feb 26, 2017
#6
+79654
0

I don't know, hectictar.....I'm not really a "Physics" person....

BTW....to be technically correct.....the answer to (3)  should be - 23.4 m/s

CPhill  Feb 26, 2017
#7
+5541
0

Oh okay, thanks

hectictar  Feb 26, 2017
edited by hectictar  Oct 17, 2017
#8
+26357
0

Constant acceleration kinetic equations can be used, as follows:

.

Alan  Feb 26, 2017

### 4 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details