Calculating Delta h for a Projectile Fired at 50 Degrees without Air Resistance

[a2k22]

Homework Statement

A cannon fires a 0.626 kg shell with initial
velocity vi = 9.6 m/s in the direction θ = 50◦
above the horizontal.

The shell’s trajectory curves downward be-
cause of gravity, so at the time t = 0.392 s the
shell is below the straight line by some verti-
cal distance Delta h. Your task is to calculate the
distance Delta h in the absence of air resistance.
What is Delta h?

Homework Equations

I have no idea.

The Attempt at a Solution

I started out by trying to by using using distance y equals initial distance y plus initial y velocity times time plus one half accel times time squared. I plugged in the numbers for where accel was 9.8 (for the parabola) and accel was 0 (for a straight line).

My teacher has not taught us this, and I'm very impressed we've gotten this far

 

[Aftermarth]

i think you are on the right track...
i don't quite understand the question but ill try give you a hand

its projectile motion so u need those equations
and you are dealing in the y direction only...

have you broken up the initial velocity?

 

[a2k22]

No, why would I need to?

 

[nasu]

I started out by trying to by using using distance y equals initial distance y plus initial y velocity times time plus one half accel times time squared. I plugged in the numbers for where accel was 9.8 (for the parabola) and accel was 0 (for a straight line).

My teacher has not taught us this, and I'm very impressed we've gotten this far

Did he/she teach you about finding the components of the initial velocity? (use the angle)
Or about writing two equations of motion, one for vertical direction (accelerated motion) and one for horizontal direction (uniform speed)?
For this problem you don't really need to look at the horizontal direction. Write eq of motion for vertical motion, once with accleration and a second time for the hypothetical case when a=0.

 

[a2k22]

Yes, I was taught about how to find v in y direction and such. What equation should I use?

 

[Aftermarth]

you will only need Vy it seems here

y = Vy*(t) - (1/2)gt^2

 

[a2k22]

Okay I got 2.13 m. Then would I do what I did before, plug it in with the accel being equal to zero, then subtract the two?

 

[Aftermarth]

this straight line... if it is the path of the shell with a=0 then yes that is right

 

[a2k22]

Can I get an answer to check mine with?

 

[Aftermarth]

okies u got to post yours first cause I am not allowed to give u the answer but :)

 

[a2k22]

ah not cool. I got 1.86.

It was 2.88 (straight line)-1.86 (parabola)
HOwever I feel I completely screwed up

 

[Aftermarth]

ok how did u get ur answer for the parabola?\
you had 2.13m above and its changeD?

 

[a2k22]

For the parabola, I set everything the same cept for the accel, which became 0, so i basically had dist= vel * time.

And I don't know where i got 2.88, and now I'm getting 3.63. Confirmation...?

 

[Aftermarth]

y = Vy*(t) - (1/2)gt^2

u have Vy i take it already
you know t
g = 9.81m/s^2
sub it in
(that is the parabola equation)

for the linear one
g = 0

 

[a2k22]

can you confirm my answer?

 

[nasu]

2.88 is when a=0
2.13 when a=g

 

[Aftermarth]

that is what i got

so now subtract the linear from the parabola

 

[a2k22]

Thank you guys so much