need physics expert

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

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.

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

We know the equation of displacement

D = v(i)*t + ½ * a *t ^2

Dy=