# Basic kinematics equations question

1. Dec 3, 2007

### Jbum

hi,

i would like to ask a physics question concerning two basic kinematics equations: d = v t and the projectile equation d = v t + 1/2 a t(squared). i was wondering on how exactly one would take these two equations and (using ONLY the symbols) prove that they should be equal. what i mean is, if these two equations were applied to a situation whereby two balls (one dropped and one projected in the horizontal direction) are launched at the same moment and from the same height, how would one prove that the equations are equal thus, showing that the two balls should, in fact, land on the ground at the same time since only the vertical path of motion is accounted for. note: one must remember that gravity acts on both balls at the same rate.

so to rephrase: how would one take d = v t (for the dropping ball) and d = v t + 1/2 a t(squared) (which is the "y" component for the projected ball in the vertical direction) and make them equal to each other, showing that t = t OR d = d. use only the symbols!

this is what i think i should do/ what i did so far: take d = v t and sub it into the other equation and solving for t... but i got stumped because i had trouble continuing. also, i tried it another way, making making d = v t and re-arranging for t and subbing that into the second equation. are these the right ideas? and will i get t = t OR d = d in the end? if not, then what's the idea that i should be looking for?

help is very much appreciated. thanks in advance.

Jb

Last edited: Dec 3, 2007
2. Dec 3, 2007

### stewartcs

You can't simply use d = vt for free-fall since it doesn't account for the acceleration due to gravity. You're comparing apples to oranges.

Here is some more info for you to read up on...

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

3. Dec 3, 2007

### Jbum

thanks for your input. i now realize my mistake in that sense. however, let me ask this: which equation, then, would represent the ball that is directly dropping (remember, there are two balls, one that is projected and one that is dropped straight down, i apologize if i was not completely clear in the previous post)?

4. Dec 3, 2007

5. Dec 3, 2007

### Jbum

love the link, but unfortunately y = y + vt + 0.5at(squared) confuses me more as i have learned y = vt + 0.5at(squared). i have never seen such an equation as the one in that link in my life. it's not that i don't believe it, it's just that i'd rather learn in terms of the equations i have been exposed to already. however, thanks again.

this is the work i have done:

v t + 1/2 a t(squared) = v t + 1/2 a t(squared)
t(v + 1/2at) = t(v + 1/2at)

after factoring the ts the "(v + 1/2at)" for both sides can be eliminated thus leaving t = t.

is this correct?

the reason that the "(v + 1/2at)" can simply cancel out is because both balls initially start at rest (where v = 0) and both balls experience the same rate of acceleration when they are "let loose". does it then make sense to cancel out "(v + 1/2at)" from both sides of the factored equation (from previous post)? but what i am not sure about is the fact that one t still remains within the brackets.. does this mean that one can just make an assumption that it is equal and thus, cancel it out? or am i wrong about this?

6. Dec 3, 2007

### Staff: Mentor

Hi Jbum,

The equation

d = v t

is just a special case of the equation

d = v t + 1/2 a t^2

where a=0. So you can always use the second, more general equation, and explicitly set a=0 whenever you have a no acceleration situation. For a projectile you have a=-g in the vertical direction and a=0 in the horizontal direction.

Last edited: Dec 3, 2007
7. Dec 3, 2007

### mich

Hi Jb:

What you did was fine,Jb. From the equation itself, we know the value of the two "t" s (the one involving an acceleration and the one involving a constant velocity) are actually the same. The ball leaving your hand, travelling in a horizontal direction, has a constant horizontal velocity (leaving air resistance aside), which will continue for a time (t) ( till the ball hits the ground) and also has a downward constant acceleration for that same period of time, of course.The downward acceleration is just like dropping the ball.

mich

8. Dec 4, 2007

### t-money

Both balls will hit the floor at the same time, because the net force in the Y-direction is equal due to -g, Therefore d=vt does not make since because it solely accounts for the X-direction, in which the ball dropped has only a Y-component.

9. Dec 6, 2007

### crowdiddly

The formula d=vt only holds true for no acceleration. It can also be modeled for average acceleration, where Vav=(Vo+Vf)/2.

So in this case, you can prove that the two formulae are equal by equating them like so:

Vav t = Vo T + 1/2 a T^2

Now, a=(Vf-Vo)/T, so...

(Vo+Vf)/2 T = Vo T + 1/2 (Vf-Vo) T
= (Vo + (Vf - Vo)/2) T
= (2 Vo + Vf - Vo)/2 T
= (Vo + Vf)/2 T

Finally, for projectile motion, the horizontal distance travelled is not necessarily equal to the vertical distance traveled, so the above derivation is useless. The two times are equal, because like you said the only force acting on an object in free-fall is gravity.

y=Vy t + 1/2 g t^2
1/2 g t^2 + Vy t - y = 0
t = -Vy +- sqrt (Vy^2 + 4 1/2 g y)/g

x=Vx t
t= x/Vx

t = -Vy +- sqrt (Vy^2 + 4 1/2 g y)/g = x/Vx

Now if you have your data, you can verify that they hit the ground at the same time.