Why Does Initial Velocity Not Affect Vertical Distance Travelled?

[Pranav Jha]

why doesn't the vertical distance traveled in the given time ,under the action of gravity alone, below the straight line it would follow in the absence of gravity, depend upon the angle of projection of a projectile?

With no gravity, the projectile, which is projected at an angle at certain initial velocity, would follow a straight-line path (dashed line). But, because of gravity, the projectile falls beneath this line the same vertical distance it would fall if it were released from rest.

I am confused about the "rest" part. why is the formula 1/2(gt^2) but not (ut+1/2(gt^2)). Why is the vertical component of initial velocity not included?

 

[K^2]

Because if you fire projectile horizontally, its initial vertical velocity is zero. Same as if you simply release it from rest.

If the projectile is fired at an angle, yes, you must include initial velocity.

 

[Pranav Jha]

With no gravity, the projectile, which is projected at an angle at certain initial velocity, would follow a straight-line path (dashed line). But, because of gravity, the projectile falls beneath this line the same vertical distance it would fall if it were released from rest.

this part explicitly refers to using the condition for rest in spite of having initial velocity in the upward direction. So, is it wrong?

 

[K^2]

this part explicitly refers to using the condition for rest in spite of having initial velocity in the upward direction. So, is it wrong?

Oh, I misread it slightly.

It's still not wrong. The dashed line is y₀ = ut. What it would move like if there was no gravity. (You can plot x or t on the x-axis, since it will only alter the scale.) The actual trajectory is y₁ = ut - (1/2)gt², as you point out. The difference between the two is y₁-y₀ = -(1/2)gt². So relative to dashed line it's moving as if it was released from rest.

Basically, it's a change of coordinate system. Instead of looking at it from perspective of someone at rest, they are looking at it from perspective of a hypothetical bullet that's unaffected by gravity. Relative to it, the fired bullet is just free falling "from rest".

 

[Pranav Jha]

hmmm... so from the reference frame of someone going up at the an initial velocity but unaffected by gravity, the other person who has the same initial velocity but is affected by gravity appears to be falling from rest.

 

[MasterX]

Newton's 2nd law is very useful for this type of problems. Also, remember that this law is written in a vector form (x,y,z-direction), and what happens in one direction does not affect the others (unless you have air-friction or other complicated phenomena happening).

Thus, for the x-direction you have
dvx/dt=0 (no force) (1)
and for the z-direction you have
dvz/dz=-g (only gravity) (2)

The first eqn gives vx=constant => x=vx*t+c1
whereas the second eqn gives vz=-g*t+c2 => z=-(1/2)g*t^2+c2*t+c3

If there was no gravity, the second eqn would have given z=c2*t+c3, a straight line

You may fire a projectile with only horizontal velocity, but, you need to fire it from a certain height, otherwise it won't go far away.

Also, if you solve the above system of eqns for the maximum distance, you will find that you need to fire the projectile at an angle of 45o.