Supposedly easy physics problem =( [projectile]

[akshajkadaveru]

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

 

[haruspex]

Looks ok so far (a and b)
What equation can you write down for the vertical motion that involves final velocity?

 

[akshajkadaveru]

v^2=v0^2+2a(x-x0) is that it ?
im really confused...

 

[haruspex]

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.

 

[akshajkadaveru]

what can i do do instead ?

 

[haruspex]

what can i do do instead ?

How does velocity change with time under a constant acceleration?

 

[akshajkadaveru]

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

 

[haruspex]

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?

 

[akshajkadaveru]

rate of change ? 1

 

[haruspex]

rate of change

Rate of change of what?

1

1 what?

 

[akshajkadaveru]

velocity ms2.. i

 

[haruspex]

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?

 

[akshajkadaveru]

dv/dt ?

 

[haruspex]

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.

 

[Bulletmittx]

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.

 

[akshajkadaveru]

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

 

[haruspex]

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.

 

[akshajkadaveru]

acceleration is 0
initial speed is 4m/s
time .5
is that correct? after that, what do i need to do ?

 

[Bulletmittx]

the time should be 0.25 seconds ._.

 

[akshajkadaveru]

who's ur teacher ?

 

[Bulletmittx]

My AP Physics teacher?

 

[haruspex]

the time should be 0.25 seconds ._.

0.5 seconds looks right to me. How did you get that?

 

[haruspex]

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?

 

[akshajkadaveru]

oh...sorry
accel = -9.8
initial = 3.46
time = .5 seconds

 

[haruspex]

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?

 

[akshajkadaveru]

-9.8(.5)+v=3.46
v=8.36

 

[haruspex]

-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.

 

[akshajkadaveru]

-1.44?

 

[haruspex]

-1.44?

Yes (but always include units).
On to part (d)?

 

[akshajkadaveru]

HOLY MOLY...in all honesty.. am i over complicating this ?

 

[haruspex]

HOLY MOLY...in all honesty.. am i over complicating this ?

Quite possibly.

 

[akshajkadaveru]

so for D... i already have the x y values and the angle right? so now i just have to plot them or something >

 

[haruspex]

so for D... i already have the x y values and the angle right? so now i just have to plot them or something >

Plot them if you wish, but to answer the question you should use algebra and calculation.
If the horizontal and vertical velocities are u and v respectively, what is the overall speed, and what is the angle of it?

 

[akshajkadaveru]

i know how to find the angle.. but how do i find the overall speed... is that by the pythagorean theorem and i plot that based on what i got for the V0x and v0y?

 

[haruspex]

overall speed... is that by the pythagorean theorem

Yes.

 

[akshajkadaveru]

2^2+3.46^2=15.97
sqrt(15.97)
3.99 m/s
tan^-1(3.46/2)= 59.97 degrees
3.99m/s @ 59.97 degrees'

 

[haruspex]

2^2+3.46^2=15.97
sqrt(15.97)
3.99 m/s
tan^-1(3.46/2)= 59.97 degrees
3.99m/s @ 59.97 degrees'

You've used the initial vertical velocity. As a result, you've recalculated the initial angle and overall speed.
Use the vertical velocity when it hits the target.

 

[akshajkadaveru]

ARGHGHGHH! so what i JUST did was wrong but what will my X and Y value things be then ?

 

[haruspex]

ARGHGHGHH! so what i JUST did was wrong but what will my X and Y value things be then ?

You posted the correct final vertical velocity in post #27. Just use that instead of 3.46 in your calculation in post #36.

 

[akshajkadaveru]

i didn't post anything on 27

 

[haruspex]

i didn't post anything on 27

I meant 28.

 

[akshajkadaveru]

so is it... 1^2+1.44^2=whatever
sqrt(whatever)= ______ @ tan^-1(1.44/1)?

will that be my answer?

 

[haruspex]

so is it... 1^2+1.44^2=whatever
sqrt(whatever)= ______ @ tan^-1(1.44/1)?

will that be my answer?

Where are you getting 1m/s from for the x direction?

 

[akshajkadaveru]

oh... i thought that't just the length of how far the thing went... how do i get that then >?

 

[haruspex]

oh... i thought that't just the length of how far the thing went... how do i get that then >?

No, you can't add a velocity to a length, or a velocity-squared to a length-squared. It makes no sense.
To apply the arctan formula and Pythagoras, you need two things of the same type at right angles - both lengths, or both velocities, or both accelerations...

What is the initial horizontal velocity? What is the horizontal acceleration? So what is the final horizontal velocity?