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Homework Statement
ball is thrown at slighta angle upwards and to the right. Angle between horizontal and the launch velocity vector is 35deg upwards.
Launch velocity vector (v_0) has magnitude of 12m/s
Only gravity affects the ball as a force. (no air resistance)
t_0 = 0 secs
t_1 = 0.2 secs
1. calculate what the velocity (##v_1##) is 0,2s after the launch.
2. work from the previous scenario and what is the displacement of the ball, ostensibly between the time period of ##t_0=0s## and ##t_1=0,2s##
Homework Equations
I tryt to use the relevant equations as I go along.
- vector components
- v_1=v_0+at
- trigonometric functions
The Attempt at a Solution
I was quite a bit confused conceptually speaking about the total displacement between the time period of 0.2s from the 0s timepoint onward.1. velocityaccording to my notes, it should be so that the ##\vec{v_1}## can be constructed from components.
it was known in the beginning that ##V_{0x}= V_{1x}## because the only force affecting is in the vertical direction from gravity. Thus I think the x direction is unaffected by a force. y direction does have gravity though.
In the y direction, we have originally ##V_{0y} = sin(35)*|\vec{V_0}|~~=6.8829m/s##
and originally in the x direction we have ##V_{0x}= 12m/s*cos(35)~~=9.8298m/s##
##V_{1y}= V_{0y}-g*Δt~~=4.9209m/s##
##V_{1y}=6.8829m/s -(9.81\frac{m}{s^2}*0.5s)##
##V_{1x}=9.8298m/s##
##V_{1y}= 4.9209m/s##
##\vec{V_1}## will be constructed from components
and the angle β between from horizontal upwards can be calculated
##tan(β)=\frac{4.9209}{9.8298}##
β=26.5930 deg## sin(26.5930) = \frac{V_1y}{|\vec{V_1}|}~~##
##|\vec{V_1}|=10.9927 m/s##
2. Displacement
this was much harder for me. I could not really remember much useful facts about the displacement formulas etc...
I did seem to grasp that what the teacher wanted us to do was as follows: essentially find the position x_1 and y_1 of the ball. Then you have found the endpoint coordinates. And we know already that the coordinates for the launchpoint x_0=0m and y_0= 0m
displacement is probably the vector from the launchpoint until the endpoint. Displacement is a vector quantity.
I think we know from the endpoint, that this position occured, at the timestamp of 0.2 seconds. (?)
I think the teacher gave us a formula for this type of ball throw scenario to use. But I'm not very confident in knowing and understanding that displacement formula and why it should work the way it does.
Rather than simply engage in a mindless exercise about plugging and chugging values into a formula. I decided to ask about my confusion about that formula. (especially the displacement formula for component form of the displacement vector)
component form equation for velocity :
##V_{x0} = V_{x1}## (((essentially it seems this is from the fact that there was no x-directed forces and no x-direction acceleration affecting the ball)))
##V_{y1} = V_{y0} - g*Δt## (((essentially to my mind this formula means that the y component of velocity will be that much smaller velocity. This is because the small g acceleration does affect indeed in the y component direction. Because small g is the acceleration due to gravity and is in the same direction as the Gravity force)))
component form equation for displacement
## X_{1}= X_{0} + \frac{V_{x0}+V_{x1}}{2}* Δt## I think that the X_1 and X_0 here probably refer to what the x coordinate are for the purpose of calculating eventually displacement. I think X_0 was essentially x value from the origin (x value from launchpoint 0meters)
##Y_1= Y_0 +\frac{V_{y0}+V_{y1}}{2}*Δt## In similar vein for Y_1 and Y_0, but I'm having a hard time justiffying the formula to myself. Do the two displacement formulas really work llike that , and why should they work like that. It was not immediately obvious to me.
1.)Apparently one source of my confusion is the justification for the [(V_0 + V_1)/2] * Δt.
Actually the entire displacemnt formulas are rather confusing.2.)Why did the formula only have the component values as the fraction terms ##\frac{V_{y0}+V_{y1}}{2}##
I think in simpler linear motion problems we just used the regular equation ##s= (1/2)*a*t^2##
where does the idea come from that:
##\frac{V_{y0}+V_{y1}}{2}*Δt= (y_1 -y_0)##
3.) I thought that the formula for average velocity was basically as follows ##\frac{Δx}{Δt}=v_{avg}##
But our teacher wrote in his notes essentially that "average velocity between t_0 and t_1 time interval is
##\frac{v_0 + v_1}{2} ##"
how could both be true for average velocity, unless they really are both identical compared to each other, of course in that case all would be well.
furthermore
##Δx = \frac{v_0 + v_1}{2}*Δt##4.) I have a vague memory that the kinematic equation ##s= 0.5* a* t^2## was somehow related to the idea of taking the average out of the velocities, and multiplying by time.
Our teacher did make a graphical proof of some sort, but I forgot what it looked like sadly. Apparently its a relatively common high school proof type for this equation. Apparently the more complicated justifications usually involve calcuulus principles, which we haven't really done so far.
I know that the kinematic equation deals with calculating total displacement in motion where acceleration is constant value.I think for me the issues 2.) and 3.) are the worst sources of confusion at the moment.
If anybody knows a graphical based proof (geometric proof) for the
##Δx = \frac{v_0 + v_1}{2}*Δt##
I would like to see that picture or video where it is explained.