Prove this equation for projectile motion

In summary, to prove the equations of projectile motion, one starts with the basic kinematic equations, applying them to the horizontal and vertical components of motion separately. The horizontal motion is uniform, since there is no acceleration, while the vertical motion is subject to gravitational acceleration. By defining initial velocity, angle of projection, and time of flight, one can derive key equations such as the range, maximum height, and time of flight, demonstrating the relationship between these variables and confirming the trajectory of the projectile as a parabolic path.
  • #1
Winner123
1
0
Homework Statement
An object launched with a speed of vi from a height deltay above the horizontal floor will land a certain horizontal distance deltax from the launch point. the initial speed can be shown to be given by
vi=√(xf-xi)^2g/2(yf-yi)
proof this equation
Relevant Equations
vi=√(xf-xi)^2g/2(yf-yi)
I tried using the formulas x=xi+vit and y=yi+voyt-1/2g(t^2)
I assumed voy would be 0 and I almost arrive to the answer but idk how to get rid of the negative
 
Physics news on Phys.org
  • #2
What did you do and where did you get stuck?
 
  • #3
Winner123 said:
Homework Statement: An object launched with a speed of vi from a height deltay above the horizontal floor will land a certain horizontal distance deltax from the launch point. the initial speed can be shown to be given by
vi=√(xf-xi)^2g/2(yf-yi)
Presumably the initial velocity is horizontal. Did you leave that out or was it omitted in the original?
The given answer is clearly wrong since ##y_f-y_i## is negative.
Please clarify the scope of the square root by using parentheses or, better, using LaTeX.
 
  • #4
Winner123 said:
##\dots## but idk how to get rid of the negative

Given that the initial velocity is horizontal, you are asked to show that the initial speed can be written as $$v_i=\sqrt{-\frac{g(x_f-x_i)^2}{2(y_f-y_i)}}.$$ If you have already arrived at this expression, there is no negative sign to get rid of because ##(y_f-y_i)<0.##
 
  • Like
Likes PeroK
  • #5
You need a degree in cryptography to be a homework helper these days. Is this something to do with the Generation Z that I've been hearing about?
 
  • #6
PeroK said:
You need a degree in cryptography to be a homework helper these days. Is this something to do with the Generation Z that I've been hearing about?
idk
 
  • Like
  • Haha
Likes gmax137 and PeroK
  • #7
kuruman said:
Given that the initial velocity is horizontal, you are asked to show that the initial speed can be written as $$v_i=\sqrt{-\frac{g(x_f-x_i)^2}{2(y_f-y_i)}}.$$ If you have already arrived at this expression, there is no negative sign to get rid of because ##(y_f-y_i)<0.##
As I posted, the given equation is wrong because it does not have that minus sign, unless your eyes are much better than mine.
 
  • #8
haruspex said:
As I posted, the given equation is wrong because it does not have that minus sign, unless your eyes are much better than mine.
Indeed it does not. In post #4 I show the equation that OP is supposed to show in readable form. We don't know whether the equation as given to OP has or does not have the minus sign. I suspect that OP got the correct equation but posted what is to be shown without it thinking that no minus sign belongs under a radical ever. Then OP did the algebra correctly, ended up with a minus sign and is asking us how to "get rid" of the minus sign.

Convoluted? Perhaps, but this is what happens when it is drilled in one's head that overall negative signs do not belong under radicals. There is an easy way to find out, so let's wait to see what OP has to say.
 
  • Like
Likes PeroK
  • #9
kuruman said:
Indeed it does not. In post #4 I show the equation that OP is supposed to show in readable form. We don't know whether the equation as given to OP has or does not have the minus sign. I suspect that OP got the correct equation but posted what is to be shown without it thinking that no minus sign belongs under a radical ever. Then OP did the algebra correctly, ended up with a minus sign and is asking us how to "get rid" of the minus sign.

Convoluted? Perhaps, but this is what happens when it is drilled in one's head that overall negative signs do not belong under radicals. There is an easy way to find out, so let's wait to see what OP has to say.
Since the OP's issue seems to be trying and failing to get rid of the minus in his/her own attempt, it seems unlikely that such a minus sign in the given equation was overlooked.
 
  • #10
Although it's rare, some textbook authors write ##g=\pm 9.8\ \mathrm{m/s^2}## with a convention for when to use which sign.
 
Last edited:
  • #11
haruspex said:
The given answer is clearly wrong since ##y_f-y_i## is negative.
Not if the positive y direction is taken to be downwards.
 
  • #12
PeroK said:
Not if the positive y direction is taken to be downwards.

Winner123 said:
from a height delta y above the horizontal floor
 

FAQ: Prove this equation for projectile motion

What is the basic equation for projectile motion?

The basic equations for projectile motion are derived from the kinematic equations. The horizontal motion is given by \( x = v_0 \cos(\theta) t \) and the vertical motion is given by \( y = v_0 \sin(\theta) t - \frac{1}{2} g t^2 \), where \( v_0 \) is the initial velocity, \( \theta \) is the angle of projection, \( t \) is the time, and \( g \) is the acceleration due to gravity.

How do you derive the time of flight for a projectile?

The time of flight \( T \) can be found by setting the vertical displacement \( y \) to zero and solving for \( t \). Using the equation \( y = v_0 \sin(\theta) t - \frac{1}{2} g t^2 \), we get \( t \left( v_0 \sin(\theta) - \frac{1}{2} g t \right) = 0 \). Solving this, we get \( t = 0 \) or \( t = \frac{2 v_0 \sin(\theta)}{g} \). The time of flight is \( T = \frac{2 v_0 \sin(\theta)}{g} \).

How can you determine the maximum height reached by a projectile?

The maximum height \( H \) is reached when the vertical component of the velocity becomes zero. Using the vertical motion equation \( v_y = v_0 \sin(\theta) - g t \) and setting \( v_y = 0 \), we get \( t = \frac{v_0 \sin(\theta)}{g} \). Substituting this back into the vertical displacement equation, we get \( H = v_0 \sin(\theta) \left( \frac{v_0 \sin(\theta)}{g} \right) - \frac{1}{2} g \left( \frac{v_0 \sin(\theta)}{g} \right)^2 \), simplifying to \( H = \frac{(v_0 \sin(\theta))^2}{2g} \).

What is the range of a projectile?

The range \( R \) is the horizontal distance traveled by the projectile when it returns to the same vertical level from which it was launched. Using the horizontal motion equation \( x = v_0 \cos(\theta) t \) and the time of flight \( T = \frac{2 v_0 \sin(\theta)}{g} \), the range is \( R = v_0 \cos(\

Similar threads

Kinematic Equations in Projectile Motion (this approach is not working)
Replies
16
Views
2K
Why does this not work? Projectile Motion problem for Dynamics Class
Replies
19
Views
2K
Differential Equation for a Pendulum
Replies
21
Views
2K
Confused about kinematic equations
Replies
20
Views
837
Equation Demonstration -- Simple Pendulum Analysis
Replies
5
Views
1K
Kinematics motion - Dancers pausing at top of their leap
2
Replies
49
Views
6K
How Do These Equations Model Projectile Motion?
Replies
3
Views
2K
Can I Assume Yi is 0m in Projectile Motion Calculations?
Replies
4
Views
1K
Initial Position Discrepancy Involving Fluid Resistance
Replies
5
Views
1K
Need Help Deriving Projectile Motion Equations
Replies
8
Views
4K

Hot Threads

Recent Insights

Back
Top