9.8 m/s^2
because of the mass of the object
It is the same (neglecting air resistance).
The horizontal velocity of the projectile (and the air resistance if known) will determine the horizontal distance traveled and the time required.
Yes. Neglecting the effects of air resistance, all objects near the surface of the earth fall with the same constant acceleration, regardless of their mass/weight.
The vertical component of a projectile's velocity is irrelevant. It can be up, down, or zero, makes no difference. As long as projectile motion lasts ... gravity is the only force on the object and you're ignoring air resistance ... its acceleration is constant, and is equal to the acceleration of gravity: 9.8 meters per second2 pointing down.
the vertical accelaration in case of a projectile is 'g'.
because of the mass of the object
It is the same (neglecting air resistance).
Zero.
Assuming negligible air resistance, the acceleration of a projectile near the Earth's surface is always the gravitational 9.81 m/sec/sec downwards, regardless of where in the trajectory the projectile is.
The horizontal velocity of the projectile (and the air resistance if known) will determine the horizontal distance traveled and the time required.
Yes. Neglecting the effects of air resistance, all objects near the surface of the earth fall with the same constant acceleration, regardless of their mass/weight.
In the usual simple treatment of projectile motion, the horizontal component of the projectile's velocity is assumed to be constant, and is equal to the magnitude of the initial (launch) velocity multiplied by the cosine of the elevation angle at the time of launch.
The vertical component of a projectile's velocity is irrelevant. It can be up, down, or zero, makes no difference. As long as projectile motion lasts ... gravity is the only force on the object and you're ignoring air resistance ... its acceleration is constant, and is equal to the acceleration of gravity: 9.8 meters per second2 pointing down.
After, and at the exact moment, the ball leaves the hand it is only accelerated by gravity if you disregard air resistance.
it depends on the gravitational force of attraction of earth and air resistance. if we are neglecting air resistance, the max.horizontal distance is according to this formulae, V0/2 * sin (2theta) where V0 is the initial velocity theta is the angle with x axis and the projection.
Because there's no horizontal force acting on it that would change its horizontal component of velocity. (In practice, that's not completely true, since the frictional 'force' of air resistance acts in any direction. But outside of air resistance, there's nothing else acting horizontally on the projectile.)