# Kinematic Equations in Projectile Motion (this approach is not working)

• GiantYoda
In summary: I cannot see where OP made the mistake of, effectively, using an equation twice. Under constant acceleration, the equation ##\Delta y = \frac{1}{2}(v_{yi}+v_{yf})\Delta t## stands alone and relates the displacement to the average velocity and the time interval over which the average is taken. It can always be used to find anyone of the three quantities when the other two are given except when the displacement and average velocity are zero.
GiantYoda
Homework Statement
If a projectile's initial vertical velocity is 12.5 m/s. What is the time it takes in the air till it reaches the same horizontal level, neglecting air resistance?
Note: I know how it can be solved, but I am asking why can't I use the following kinematic equations?
Δy=[(vyi+vyf)/2]*Δt
Relevant Equations
Kinematic equations
Givens:
Vyi=12.5 m/s
Vyf=-12.5 m/s (at the same horizontal level)
ay=-9.81 m/s^2
Δy= zero m (as the displacement on the y-axis, when the projectile reaches the same horizontal level, is zero m)
Δt=?

When I use
Δy=[(vyi+vyf)/2]*Δt

I get the time as undefined.
Δt= 2Δy/(vyi+vyf)
= 2*0 m/(12.5 m/s +-12.5 m/s)

Any reason why can't I use this specific kinematic equation?

GiantYoda said:
Any reason why can't I use this specific kinematic equation?
You mean a reason other than it doesn't give you ##\Delta t##?

Note that the initial and final velocities and the final displacement are independent of ##g## and ##\Delta t##.

PeroK said:
You mean a reason other than it doesn't give you ##\Delta t##?
Exactly. According to what I have studied, all the kinematic equations are useable interchangeably as long as the givens allow. Why does this equation yield 0/0?

GiantYoda said:
Exactly. According to what I have studied, all the kinematic equations are useable interchangeably as long as the givens allow. Why does this equation yield 0/0?
PeroK said:
Note that the initial and final velocities and the final displacement are independent of ##g## and ##\Delta t##.

I am not sure if I got your second point. All what I am asking is since all kinematic equations are correct, all should give the same solution. Why isn't this applicable with Δy=[(vyi+vyf)/2]*Δt?

GiantYoda said:
I am not sure if I got your second point. All what I am asking is since all kinematic equations are correct, all should give the same solution. Why isn't this applicable with Δy=[(vyi+vyf)/2]*Δt?
Because in this case ##\Delta y = 0## and ##v_{avg} = 0##. You've simply chosen the wrong approach to the problem.

When you move on to more difficult problems finding the right approach is critical. Not every approach is guaranteed to work.

Lnewqban
Welcome!
The way I see it:
These equations are meant to be used between an initial and a final point of displacement.
You know that the velocity at the top of the vertical trajectory must be zero.
I would then calculate the time consumed in the upwards or downwards trajectory only; then, multiply that time by two.

Marc Rindermann
GiantYoda said:
I am not sure if I got your second point. All what I am asking is since all kinematic equations are correct, all should give the same solution. Why isn't this applicable with Δy=[(vyi+vyf)/2]*Δt?
The equation says that when the projectile is at a given height going up and at time ##\Delta t## later returns to the same height going down, the vertical component of the velocity has the same absolute value but different sign. This is true for all given heights, all initial velocities and all gravitational accelerations, i.e. whether the projectile motion takes place on Earth, the Moon, Mars etc.

So for the particular problem you treat, you know that ##\Delta y = 0## because you have stipulated that the projectile returns to the same height. You also know that the projectile needs a certain amount of non-zero time ##\Delta t## to come back down to the same height. So when you use this equation you start from $$0=\frac{1}{2}(v_{yi}+v_{yf})\Delta t.$$ We have already established that ##\Delta t## is not zero and ##\frac{1}{2}## is also not zero (not even for small values of ##\frac{1}{2}## ), so the sum ##(v_{yi}+v_{yf})## must be zero, which is as it should be for the equation to hold and as you have found out.

The problem is not that the equation does not work. The problem is with the way you interpreted it. Equations are descriptions of physical reality; they are not mathematical cranks that you turn so that you put numbers in and get numbers out that correspond to symbols appearing in them.

Angetaire, pop_ianosd, Marc Rindermann and 1 other person
GiantYoda said:
I am not sure if I got your second point.
@PeroK's point is that if ##v_{yi}+v_{yf}=0## then ##\Delta y=0##, regardless of the value of ##\Delta t##. You could double both velocities and again the displacement will be zero but the time will have increased. So you have not used the magnitude of the velocities.

With any system of equations, there may be enough to provide a solution, but if you make the mistake of, effectively, using one twice instead of using them all once you tend to end up with a tautology. Indeed, you got 0=0.

Last edited:
SammyS, Lnewqban and PeroK
haruspex said:
With any system of equations, there may be enough to provide a solution, but if you make the mistake of, effectively, using one twice instead of using them all once you tend to end up with a tautology. Indeed, you got 0=0.
I cannot see where OP made the mistake of, effectively, using an equation twice. Under constant acceleration, the equation ##\Delta y = \frac{1}{2}(v_{yi}+v_{yf})\Delta t## stands alone and relates the displacement to the average velocity and the time interval over which the average is taken. It can always be used to find anyone of the three quantities when the other two are given except when the displacement and average velocity are zero and ##\Delta t## is not. I think it is this exception that OP did not take into consideration. To me it looks like OP unknowingly started with 0 = 0 and was surprised in #4 when he ended up with 0/0.

GiantYoda said:
ay=-9.81 m/s^2
Is there a reason that you chose an equation that does not make use of this bit of information?

PeroK
jbriggs444 said:
Is there a reason that you chose an equation that does not make use of this bit of information?
I think the reason is OP's perceived interchangeability of the kinematic equations (as indicated in #4) which led to the question "why does this equation not work?"

kuruman said:
I cannot see where OP made the mistake of, effectively, using an equation twice.
How was ##|v_i|=|v_f|## obtained? It is listed in post #1 as a "given", but it is not. Presumably it is from ##v_f^2-v_i^2=2a\Delta y## and ##\Delta y=0##.
Then ##\Delta y=0## is used again in ##Δy=[(v_{yi}+v_{yf})/2]*Δt##.

haruspex said:
How was ##|v_i|=|v_f|## obtained?
I would say that OP tried to apply the average velocity equation without really realizing how that would work out in the end. It is given that that the projectile returns to the same height which means that ##\Delta y=0##. If that is substituted in the equation, one gets ##0=\frac{1}{2}(v_{yi}+v_{yf})*\Delta t##. No second equation is needed to deduce that this is satisfied if ##v_{yf}=-v_{yi}## when ##\Delta t >0##. That's where, I think, OP got confused about the use of the equation because he realized that it resulted in 0 = 0 and no solution for ##\Delta t## could be found using it.

kuruman said:
I would say that OP tried to apply the average velocity equation without really realizing how that would work out in the end. It is given that that the projectile returns to the same height which means that ##\Delta y=0##. If that is substituted in the equation, one gets ##0=\frac{1}{2}(v_{yi}+v_{yf})*\Delta t##. No second equation is needed to deduce that this is satisfied if ##v_{yf}=-v_{yi}## when ##\Delta t >0##. That's where, I think, OP got confused about the use of the equation because he realized that it resulted in 0 = 0 and no solution for ##\Delta t## could be found using it.
That doesn't answer my question. The OP's first step was to assert ##|v_i|=|v_f|##. How did the OP determine that? Only by using ##\Delta y=0##, right? Using it a second time is the trap I described.

OK, I see what you mean. The fallacy I saw was having ##A*B=0## and expecting to find ##B## when it has already been established that ##A=0##.

## 1. What are the kinematic equations used in projectile motion?

The four kinematic equations used in projectile motion are:
1. v = u + at
2. s = ut + 1/2at^2
3. v^2 = u^2 + 2as
4. s = (u + v)/2 * t
where v is the final velocity, u is the initial velocity, a is the acceleration, s is the displacement, and t is the time.

## 2. How do you apply the kinematic equations to solve projectile motion problems?

To apply the kinematic equations to solve projectile motion problems, you first need to identify the known and unknown variables. Then, you can plug in the known values into the appropriate equation and solve for the unknown variable. It is important to pay attention to the units and use consistent units throughout the problem.

## 3. What is the difference between horizontal and vertical components of projectile motion?

The horizontal component of projectile motion refers to the motion of the object in the x-axis, while the vertical component refers to the motion in the y-axis. The horizontal component is affected by the initial velocity and the acceleration due to gravity is zero, while the vertical component is affected by the acceleration due to gravity and the initial velocity in the y-direction.

## 4. Can the kinematic equations be used for any type of projectile motion?

Yes, the kinematic equations can be used for any type of projectile motion, as long as the acceleration remains constant. This means that the equations can be applied to objects thrown horizontally, at an angle, or dropped from a height.

## 5. How do you determine the range of a projectile using the kinematic equations?

The range of a projectile can be determined by using the equation:
R = (v^2 * sin2θ)/g
where R is the range, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. This equation takes into account both the horizontal and vertical components of the projectile's motion.

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