Supposedly easy physics problem =( [projectile]

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SUMMARY

The discussion revolves around solving a projectile motion problem involving an initial velocity of 4 m/s at a 60-degree angle, targeting a distance of 1.0 m horizontally. Key calculations include determining the y-coordinate of the target, which was found to be approximately 3.2 meters, and the time to hit the target, which was debated between 0.25 seconds and 0.5 seconds. The final vertical velocity was calculated to be -1.44 m/s, and the overall speed before impact was derived using the Pythagorean theorem, resulting in a magnitude of 3.99 m/s at an angle of 59.97 degrees. The discussion highlights common misconceptions and emphasizes the importance of using correct initial conditions and equations in projectile motion.

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  • Understanding of basic kinematics and projectile motion.
  • Familiarity with the equations of motion, including v = v0 + at.
  • Knowledge of trigonometric functions, particularly sine and cosine.
  • Ability to apply the Pythagorean theorem in physics contexts.
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  • Study the derivation and application of the equations of motion in projectile problems.
  • Learn how to analyze projectile motion using vector components.
  • Explore the effects of varying launch angles on projectile trajectories.
  • Investigate the impact of air resistance on projectile motion calculations.
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Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators looking for examples of common student misconceptions in these topics.

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You launch a projectile with an initial velocity of 4 m/s at an angle of 60 degrees and it hits a target at a distance of 1.0 m in the horizontal direction.
a. what is the y coordinate of the target
b.how long does it take for the projectile to hit the target
c.what is the y- component of its velocity just before it impacts the target
d. and what is its velocity ... magnitude and idrection, just before it impacts the target ?this is what i did

1. drew graph
2.cos60=vox/4m/s
vox=2m/s ; ax=0
3.sin60= voy/4m/s
voy= 3.46 m/s ; ay = -9.8ms2
4.time to travel and hit the target.
1.0/2 = .5 seconds
5. y coordinate you can find the equation with
y=voyT+1/2ayT^2
plugging in what i know, i got .51

im REALLY really confused.. please enlighten me
 
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Looks ok so far (a and b)
What equation can you write down for the vertical motion that involves final velocity?
 
v^2=v0^2+2a(x-x0) is that it ?
im really confused...
 
akshajkadaveru said:
v^2=v0^2+2a(x-x0) is that it ?
im really confused...
It will do, though it is not ideal here. It uses the final height (x) determined earlier, so plugging in the numerical value for that leads to some accumulation of rounding errors. It would be better to use an equation with initial velocity, acceleration and time. Yes, time is also a calculated value, but it was determined precisely.
 
what can i do do instead ?
 
akshajkadaveru said:
what can i do do instead ?
How does velocity change with time under a constant acceleration?
 
im sorry but I am not really catching on... can you explain your thought process so therefore i can understand the other previous problem better... i know this question should be REALLY easy... i feel like I am missing out on something that is really obvious
 
akshajkadaveru said:
im sorry but I am not really catching on... can you explain your thought process so therefore i can understand the other previous problem better... i know this question should be REALLY easy... i feel like I am missing out on something that is really obvious
Let's start with the basics. What is acceleration? If a car goes from stationary to 20 m/s in 10 seconds, what is its average acceleration?
 
rate of change ? 1
 
  • #10
akshajkadaveru said:
rate of change
Rate of change of what?
akshajkadaveru said:
1
1 what?
 
  • #11
velocity ms2.. i
 
  • #12
akshajkadaveru said:
velocity ms2.. i
Yes, acceleration is the rate of change of velocity. Put that in an equation. If the velocity changes by ##\Delta v## in time ##\Delta t##, what is the average acceleration?
 
  • #13
dv/dt ?
 
  • #14
akshajkadaveru said:
dv/dt ?
OK, but that notation is usually reserved for continuous changes, dv/dt being the instantaneous acceleration. Here we are concerned with a change in speed over some significant period of time, so we use the ##\Delta## notation instead.
So we have ##a = \frac{\Delta v}{\Delta t}##.
Rearrange that in the form final speed = initial speed plus something.
 
  • #15
Is it a coincidence that I have the same physics homework as you? I don't know, but so far you're wrong. Your first mistake is the time. Yours is for 2m/s not 4m/s; this means that the time it should take to travel one meter is .25 seconds not .5
This being said, I ended up with a y coordinate of 3.2 meters.
 
  • #16
i really appreciate you helping me but i don't understand how this is going to help me for the problem...
a=dv/dt
aDt=(v-v0)
adT-v=-v0

-aDt+v=v0
 
  • #17
akshajkadaveru said:
i really appreciate you helping me but i don't understand how this is going to help me for the problem...
a=dv/dt
aDt=(v-v0)
adT-v=-v0

-aDt+v=v0
Good.
In the vertical direction, you know the acceleration, the time to reach the target, and the initial speed. Plug those into the equation above.
 
  • #18
acceleration is 0
initial speed is 4m/s
time .5
is that correct? after that, what do i need to do ?
 
  • #19
the time should be 0.25 seconds ._.
 
  • #20
who's ur teacher ?
 
  • #21
My AP Physics teacher?
 
  • #22
Bulletmittx said:
the time should be 0.25 seconds ._.
0.5 seconds looks right to me. How did you get that?
 
  • #23
akshajkadaveru said:
acceleration is 0
initial speed is 4m/s
time .5
is that correct? after that, what do i need to do ?
No, vertical direction. What is the initial speed in the vertical direction? What is the acceleration in the vertical direction?
 
  • #24
oh...sorry
accel = -9.8
initial = 3.46
time = .5 seconds
 
  • #25
akshajkadaveru said:
oh...sorry
accel = -9.8
initial = 3.46
time = .5 seconds
Good, so what does your equation in post #17 give you for the final vertical velocity?
 
  • #26
-9.8(.5)+v=3.46
v=8.36
 
  • #27
akshajkadaveru said:
-9.8(.5)+v=3.46
v=8.36
Almost, but you've confused yourself with the minus signs.
Your equation in post #17 was equivalent to ##v = v_0 + a\Delta t##.
Plug your v0 and a (= -9.8) into that.
 
  • #28
-1.44?
 
  • #29
akshajkadaveru said:
-1.44?
Yes (but always include units).
On to part (d)?
 
  • #30
HOLY MOLY...in all honesty.. am i over complicating this ?
 

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