- When object is tossed in the air, nothing affects its acceleration but the acceleration of gravity only. So, the source of the acceleration is the force of gravity solely.

The objects interacting with the ball is the hand that gives force to the ball to go upward. Then as the ball is on the air, the force of gravity interacts with the ball that causes the ball to comes back to the ground and not move upward forever.

When a ball is tossed in the air, its initial velocity is zero

When given a force, it gains a velocity and go upward. The

object accelerates under the force of gravity .

here the objects interacting with the ball are the hand which gives the thrust and the gravitional force, due to which it comes back to the ground

- Supposedly an object in this motion should have not a horizontal acceleration because only the force of gravity acts on the object. So if only gravity acts on the object, only vertical acceleration is present. The possible source of the horizontal acceleration is an external force that affects the motion of the object, disabling it to accelerate in vertical only.

The gravitional acceleration = -9.8 m/s^2

Let the ball is going upward with the velocity of u m/sec e

Velocity through horizontal = v*cos x.

X is angle of the throw.

So ,

The observed acceleration in horizontal is zero.

The vertical acceleration is = -9.8 m/sec^2

The objects interacting here are the hand and the force of gravity.

- Yes the source of acceleration observed is the same with the answer in 1a. As you can see in the data the acceleration is almost close to the constant acceleration due to gravity which 9.8 m/s
^{2}.

When an object is dropped, it gains acceleration of the force of gravity.

So the height its falls its velocity Is zero

And acceleration is 9.8 m/s^2

For a tossed object

V = 3 m /sec^2. Suppose

So , when going upwards it speed = V = 3 -9.8 * t .

T is t time,

At when v= 0 , its drops due to force of gravity

So it falls with positive acceleration = 9.8 m/sec^2

So both are different

Data sheet 4

2.a

We know the equation of displacement

D = v(i)*t + ½ * a *t ^2

Dy= v(i)_{y*t }+ ½* a_{y}***t^2.**

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**Where d**_{y= }dx

2.b)

We are doing average of three columns.

K m n average

9.5755 9.6245 10.154+/- 1.57135 9.78+/-1.51715

Yes, all the data makes since.no need to eliminate one .

A_{average}+/-(est.uncertainty)=10.154+/-1.57135m/s^{2}

**2.c 7.757+/-4.556m/s**^{2}= a_{average}

**2d. ****No. The motion of the object should be always perpendicular whatever the forward horizontal velocity is. It is always independent with each other.**** Wherein by the equation of motion:**

**V**_{x}=V_{0x} ; constant all the time.

**V**_{y}=V_{oy}+at; where velocity changes uniformly by 9.8m/s^{2 }per second.

**So, example if at t-1s, and initial velocity v**_{0}y=2m/s:

**Vy=2+(-10.39)(1s)=-8.39m/s**

**At t=2s;**

**Vy=2+(-10.39)(2s)=-18.78m/s**

**So, clearly the vertical motion of the object will not be any affected by whatever horizontal motion is. It will always only be affected by the force of gravity.**

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**2e. ****The mass of the object does not in any way affect the vertical acceleration of the system. The only possibility that causes its variation is the drag force since the system is not done in a vacuum. There would be some external forces dragging the motion horizontally.**** From the equation of motion also below, mass m has nothing to do with or it’s not included, meaning it is always independent of the mass of the object. **

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**3a. ****Yes, same as the graph my data also has negative slope for both tossed and drop and almost same because as we all know the slope of a velocity-time graph is always the average acceleration of the object. Our acceleration is almost near to 9.8 m/s**^{2 }so for both tossed and drop data, so it is expected that we can same or almost same slope near to that constant value.

**The velocity becomes relatively large negative as the object moves downward approaching to the ground and it becomes relatively large positive when it moves upward. Upward motion is positive velocity that increases uniformly with the acceleration and downward motion is negative.**

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**3b. ****Yes, it will always stop at the top and velocity becomes zero. The moment it stop and velocity becomes zero at an instant where it already reaches its maximum height. This is due to the force of gravity that pulls the object downward. Without the gravitational pull any object suspended in air will continue moving in its direction of motion forever.**** **

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**3c****. the data points on the table is not very clear but you can do it the exact one I will just give you my approximation:**

**Drawing a horizontal line halfway from the minimum to maximum height meaning you are to draw from the graph from the center of the complete parabola then compute the time as:**

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**∆t(above half height) =0.66s;**** ∆t(below half height) =0.99s-0.66s=0.33s**

**∆t(total) =0.66s+0.33s=0.99s**

**The ball spends more time above half height because the height is bigger compare to the below that will affect the time of the motion of the object.**** **

** Source: www.acemyhw.com**

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