Projectile Qu Help =] (1 Viewer)

[lisarh]

A projectile is launched with a velocity of 200m/s at an angle of 45° to the horizontal. Calculate:

a. the maximum height reached:

v²= u² + 2aΔy
0 = 200² + 2(-9.8).Δy
-200² = -19.6Δy
Δy = 2040.8m
Δy = 2040.8/2
Δy = 1020 m

b. the time of flight:

?



ANS:29s



c. the range:



?


ANS: 4081m

 

[Pwnage101]

um - NO, NO , NO, NO, NO!!!!!!

(hyperventalates)

your working out is wrong, specifically i don't understand why you halved 2040 - in any case, you got the correct answer, but that's because the angle is 45 degrees - i'll show you what you did wrong:

read the equations they give you on the formula sheet for HSC - it says v_y²= u_y² + 2aΔy (note that galileo stated that the vertical and horizontal motions of a projectile act similtaneously, yet independently from one another, so we use this when we calculate these typs of questions)

now u_y = 200sin45 = 141.4s m/s [=100root2 exactly] (here we are finding the vertical component of the velocity, to use in the equation, and does not = 200 as you subbed in)

now a)

v_y²= u_y² + 2aΔy
0 = (100root2)² + 2(-9.8).Δy
0 = 20 000 + 2(-9.8).Δy
(19.6)Δy= 20 000
Δy= 20 000/19.6
1020.41
Δy=1020m

(b)
v_y= u_y + a_yt
time of flight = 2x time until maximum height reached = 2x t_0.5

v_y= u_y + a_yt
0= 141.4 + (-9.8)t_0.5
t_0.5 = -141.4/-9.8
t_0.5 = 14.43 s

thus, time of flight = 2xt_0.5= 2x 14.43 = 28.86s = 29s

(c)
another formula from the sheet:

Δx=u_xt

now u_x = 200cos45=141.4 (u_y = u_x because the angle is 45 degrees)

Δx=141.4x28.86 = 4080.8 = 4081m

[NOTE: This is a good example of where using the full solution to a previous part is paramount when using that data in a latter question, for example if i took the answer from (b) as 29 seconds, then i would calculate Δx=141.4x29 = 4 100.6, which is significantly off the actual answer, but since i took it to 2 decimal points, it was more accurate]

 

[4theHSC]

A projectile is launched with a velocity of 200m/s at an angle of 45° to the horizontal. Calculate:

a. the maximum height reached:

v²= u² + 2aΔy
0 = 200² + 2(-9.8).Δy
-200² = -19.6Δy
Δy = 2040.8m
Δy = 2040.8/2
Δy = 1020 m

b. the time of flight:

?



ANS:29s



c. the range:



?


ANS: 4081m

Ok it is always helpful to draw a diagram in projectile questions...

I have found it very helpful to always find the initial horizontal and vertical velocity, because even though they don't ask it in some questions, you have to derive it yourself so that you can get all the other working out.

Now:
Uy= 200sin45 = 141.4 m/s
Ux= 200cos45 = 141.4 m/s
To find the time of flight:
Vy= Uy +at
0= 141.4+ (-9.8)t
141.4/9.8= 14.4
But because we chose the point Vy=0 (halfway) we only calculated the time half way, so to get the full time we multiply by 2.
14.4x2= 28.86= 29 seconds (note that don’t round up your answers in exams, meaning don’t go to the nearest second)
Now we calculate the maximum height:
We know that at V=o it is the maximum height the projectile reaches, therefore, we use the quation V2y= Uy2 +2delta Y
02= (141.4)2 +2(-9.8)deltaY
Therefore: (141.4)2/19.6= deltaY
Therefore deltaY which is the maximum vertical height= 1020.1 m
Now we have all the information we need to calculate the horizontal displacement which is delta X.
deltaX= UxT
initial horizontal velocity= 141.4m/s, and the time is 28.86 (use your unrounded answer all the time so that the values are not altered)
therefore: 141.4x 28.86= 4080.8m= 4081m
hope that helps , I’m doing projectile motion questions at the moment as well

 

[4theHSC]

um - NO, NO , NO, NO, NO!!!!!!

(hyperventalates)

your working out is wrong, specifically i don't understand why you halved 2040 - in any case, you got the correct answer, but that's because the angle is 45 degrees - i'll show you what you did wrong:

read the equations they give you on the formula sheet for HSC - it says v_y²= u_y² + 2aΔy (note that galileo stated that the vertical and horizontal motions of a projectile act similtaneously, yet independently from one another, so we use this when we calculate these typs of questions)

now u_y = 200sin45 = 141.4s m/s [=100root2 exactly] (here we are finding the vertical component of the velocity, to use in the equation, and does not = 200 as you subbed in)

now a)

v_y²= u_y² + 2aΔy
0 = (100root2)² + 2(-9.8).Δy
0 = 20 000 + 2(-9.8).Δy
(19.6)Δy= 20 000
Δy= 20 000/19.6
1020.41
Δy=1020m

(b)
v_y= u_y + a_yt
time of flight = 2x time until maximum height reached = 2x t_0.5

v_y= u_y + a_yt
0= 141.4 + (-9.8)t_0.5
t_0.5 = -141.4/-9.8
t_0.5 = 14.43 s

thus, time of flight = 2xt_0.5= 2x 14.43 = 28.86s = 29s

(c)
another formula from the sheet:

Δx=u_xt

now u_x = 200cos45=141.4 (u_y = u_x because the angle is 45 degrees)

Δx=141.4x28.86 = 4080.8 = 4081m

[NOTE: This is a good example of where using the full solution to a previous part is paramount when using that data in a latter question, for example if i took the answer from (b) as 29 seconds, then i would calculate Δx=141.4x29 = 4 100.6, which is significantly off the actual answer, but since i took it to 2 decimal points, it was more accurate]

LOL nice

I have a question: when they ask you to analyse projectile motion diagrams (Page 13 of surfies book space if you have it) and they tell you something like the frequency of the stroboscope is 10Hz. What does that mean?

 

[scora]

jesus pwnage that must've taken you like 20 mins to type that thing out

 

[Pwnage101]

4theHSC, 10HZ as a frequency means basically they took a photo 10 times a second, and stroboscope means they overlaid the pictures, so basically, if it's to scale u can find out the distance travelled, and you haave the corresponding time between each photo/position (ie0.1 seconds), from there u can calculate other stuff

scora, yeh well i try my best to help fellow BOSer's out =]

 

[darkchild69]

Geezus, every bloody time i come on to see if questions need to be answered, you have already answered them!

FFS!

 

[Pwnage101]

sorry, darkchild69

but i'm sure you are better qualified to asnwer the questions, so if you see a mistake or just want to add your 2 cents to the mix, feel free

 

[lisarh]

Geezus, every bloody time i come on to see if questions need to be answered, you have already answered them!

FFS!

It's ok, I'm pretty sure I'm going to have plenty more questions to ask. x]

 

[cutemouse]

A projectile is launched with a velocity of 200m/s at an angle of 45° to the horizontal. Calculate:

a. the maximum height reached:

S=(u²sin²2θ)/2g
=(200²sin²45)/(2*9.8)
=1020.41m

b. the time of flight:

t=(2usinθ)/g
=(2*200*sin45)/9.8
=28.86s

c. the range:

Δx=(u²sin2θ)/g
=(200²*sin90)/9.8
=4081.63m