please explain what are the values of the verical and horizontal accerlrations of a projectile?

Take your typical rifle say the AK 47.  Then take your typical angle finder say the craftsman from Big Lots, and attach it to the barrel of your assault rifle using the magnet attached to the bottom.  Then assume no gravity.

1)hold the gun so that the angle reads 0 and fire it.  the bullet travels with 100% horizontal velocity and 0% vertical. expressed mathmatically (trigonometry) Sin = 0, Cos=1

2) repeat with the angle reading 90 and fire again.  this time the bullet has 0% horizontal and 100% vertical velocity.  Sin =1, cos =0.

At any other elevation, there is a combination of vertical and horizontal components as the sine and cosine values lie between 0 &1.  By multiplying the velocity by the sine for vertical component and the cosine for the horizontal, both components can be figured out.

It is a basic right triangle solution.  Vertical component is the rise and the horizontal is the run.  The angle + the speed/velocity is called a vector and is indicated by an arrow at the angle of the barrel aka the angle of inclination .

it gets a bit more complicated when we return to gravity because it imposes a downward component of -32 feet  second.  it is negative because it pulls the bullet down against the upward component of the bullet.  of course if you shot down, then the vertical part would also be negative and gravity would increase the vertical component.

hope this is understandable.

Think how a projectile is moving. It has upward and lateral displacements. To move up it should have certain velocity in the upward direction. This part of its velocity is called the vertical velocity. If Q is the angle made by the total velocity u with the horizontal, then
vertical velocity = usinQ
To move laterally the projectile, it should have a velocity component in the that direction. We usually take this in a direction parallel to the ground and it is called the horizontal velocity.
horizontal velocity = ucosQ
Resultant velocity, u = square root of(hor. velocity^2 + vert. velocity^2).
For more details. contact me. My id is rajsruthilayam@yahoo.co.in

Dearest B_Ball
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Just an example for u OK !
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Physics/Math Question?
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The launching velocity of a projectile is 20 m/s at 53 degrees above the horizontal.

A. What is the vertical component of its velocity at launch?

B. Its horizontal component of velocity?

C. Neglecting air friction, which of these components remains constant throughout the flight path?

D. Which of these components determines the projectile's time in the air?

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just for reference
g=32.2ft/s²
r=unknown

?=0.5 * ((invsin)(r*g/v.²))
2?=((invsin)(r*g/v.²))
2?*v.²=((invsin)(r*g))
sin (2?*v.²)=r*g

Vy=15.96
Vx=12.036

therefore
sin (2?*v.²)/g=r
and
sin (2?*v.²)/r=g
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first thing u do is look at the question
of course
u have 2 angles
1./ 53º
and
2./ 90º-53º=37º
you are given the resultant speed(hyp)
so now u have to find
the compotent speeds
there are 2
speed(Y)
and
speed(X)
so lets find both
53º above the horizon
therefore

20 * cos(53º)=speed(X) or 20 * sin(37º)=speed(X)

20 * sin(53º)=speed(Y) or 20 * cos(37º)=speed(Y)

so

20 * cos(53º)=speed(X) or 20 * sin(37º)=speed(X)
20 * 0.6018 =12.036

20 * sin(53º)=speed(Y) or 20 * sin(53º)=speed(Y)
20 * 0.798 = 15.96
now we have speed(Y) = 15.96
v= g * t
v = 32.2 * 0.4956
v = 32.2 * 0.4956
v =15.96 or speed(Y)

lets find distance to drop from the sky to ground

d(Y) = 0.5 * g * t²
d(Y) = 0.5 * 32.2 * (.4956)²
d(Y) = 7.97ft or distance(Y)
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now u can calculate the Range
----------------------------
d(X) = speed(X) * t
d(X) = 12.036 * .4956
d(X) = 5.965 ft

Carry on from here now
if u cant see this now
take up social science or politics