Physics Help Please: Projectile Motion (1 Viewer)

[clissold]

Hey Everyone,

During these holidays our teacher has asked us to complete some questions from the "Dot Point HSC Physics" book. I must admit, it has been a while since Ive even looked at Physics and was just wondering if anyone could help me answering the following questions about Projectile Motion.

2.3.1 A projectile is fired horizontally at 150m/s from the top of a 196m high cliff. Calculate:

a) its time of flight
b) its range
c) its velocity on hitting the ground

2.3.2 A projectile has a time of flight of 7.5s and a range of 1200m. Calculate:

a) its horizontal velocity
b) its maximum height
c) the velocity with which it is projected

2.3.3 A cannonball is fired at 80m/s at an angle of 45 degrees to the horizontal. Calculate the height at which the ball hits a vertical cliff 150m away

Thankyou.



P.S.. Yes I do realise that the answers are in the back of the book but I need worked solutions

 

[Bored Of Fail 2]

ok

2.3 a

draw pic, take down as positiv direction

so vertical displacement = s = +196m
vertical acceleration = +g = +9.8 m/s^2
and because its projected horizontally, its initial vertical velocity= v (y) = 0


now the eqn which contains all these is s= ut + (1/2) a t^2

now sub in the numbers 196 = 0 + (4.9) t^2

so t= +- sqrt ( 196 / 4.9) , and take the + case, as time is positive , so t= 6.3sec.

b. Now we know that speed= distance/ time , so distance = speed x time

and remember the horizontal velocity is CONSTANT

so range = 150 x 6.3 = 945m

c. The velocity when it hits the ground is equal to sqrt ( ( x velocity)^2 + (y velocity)^2 )

when it hits the ground, x velocity = 150m/s, its constant

the y velocity can be found from the eqn v(y) = u(y) + a(y) t = 0 + ( 9.8) ( 6.3) = 61.7m/s

so the magnitude of the velocity = sqrt ( (150)^2 + ( 61.7)^2 ) = 163m/s

and you should find the angle with the horizontal as well, which can be done by drawing a pic

we get that x = tan^-1 ( ( y velocity ) / (x velocity) ) and its below the horizontal

so

 

[Bored Of Fail 2]

2.3.2

a.

range = horizontal speed x time

so horizontal velocity = range/time = 1200/7.5 = 160m/s

 

[Bored Of Fail 2]

someone can do the rest, i dont feel like doing anymore

 

[PPTR]

For 2.3.2: Given that: t = 7.5s and Range = 1200m.

First u want to draw a simple sketch of a ball being launched with an initial velocity (u) at an angle : theta. Also indicate the trajectory of the ball showing the maximum height and the range of the projectile.

a) Horizontal component = Range/time of flight = 1200/7.5

b)maximum height:

To find the maximum height, first recall the assumption of projectile motion where the the time taken to go up is the same as the time take to come down. By that assumption then, the time taken to reach maximum height is t/2 = 7.5/2 = 3.75s

Next using the eqn v.y = u.y + at: where 'v.y' is the final vertical velocity which is equal to zero at max. height

'u.y' is the initial vertical component.

't' is the time taken to reach max. height which is equal to 3.75s; and 'a' is 9.8 and downwards is considered positive.

Substitute the values: 0 = u.y + (-9.8)(3.75); Therefore, u.y = 36.75 m/s

Substitute 36.75 into the eqn r = (u*t) + (1/2)*a*(t^2); that should give you the maximum height.

c) From parts a and b, you have calculated the horizontal and vertical components. Find the initial velocity and the angle using simple trignometry.

Hope that was helpful...


Need Help??? Check this out....
http://community.boredofstudies.org/showthread.php?t=254227

 

[Bored Of Fail 2]

first recall the assumption of projectile motion where the the time taken to go up is the same as the time take to come down.

also remember , that it is only true if the projectile moves in a symmetric parabolic path , I dont think there is any other way of doing this question without it moving in a symmetrical parabola , but the questions really should specify that the path is symmetrical, or that it lands in the same plane that it was fired from, if it was fried from say 10m above the ground and it landed on the ground it would be obvious that you cannot just halve the time of flight to find the time when it was at a maxium height

 

[PPTR]

also remember , that it is only true if the projectile moves in a symmetric parabolic path , I dont think there is any other way of doing this question without it moving in a symmetrical parabola , but the questions really should specify that the path is symmetrical, or that it lands in the same plane that it was fired from, if it was fried from say 10m above the ground and it landed on the ground it would be obvious that you cannot just halve the time of flight to find the time when it was at a maxium height

In projectile motion for HSC, you do not consider aerodynamic drag for any questions and therefore the path remains perfectly parabolic. So it isn't necessary for them to say whether the path is parabolic or not. Also, nowhere in the question does it state its fired from a cliff so based on the information provided we can safely assume that it is fired on a level plane.

 

[clissold]

Thankyou both "Bored of Fail 2" and "PPTR"

However I still have a few questions..

Bored of Fail 2, why is it that we take down to be the positive direction? I havnt really heard of doing this although it did prove useful for this question. Also, I hope sometime in the near future you can "feel like" answering the remaining questions for me.

PPTR, could you please re-explain questions 2.3.2 b and 2.3.2 c as I dont completely understand your working. Also, thankyou for the tutoring offer although I dont live in your area sorry.

 

[Bored Of Fail 2]

http://img811.imageshack.us/f/projectile.jpg/

for last question

 

[clissold]

Okay, a few things..

To start with, I actually really appreciate the answers given by both "Bored of Fail 2" and "PPTR". Judging by their responses, they are both talented physics students and I am grateful for their assistance. However, I do not completely understand the working given by "PPTR" for questions 2.3.2 b and 2.3.2 c and was just simply asking for a further explanation (as I believe most people do when they dont completely understand something).

Furthermore, I created this thread with the intention of receiving a little bit of help with the above questions. Nowhere is there any mention of "please abuse me for not being the best physics student there ever was" so may I suggest you stop being the smartass you are coming across as and actually help me with these questions if you have any relevant physics knowledge. If not, shut the fuck up and find something better to do.

 

[Triple777ER]

Okay, a few things..

To start with, I actually really appreciate the answers given by both "Bored of Fail 2" and "PPTR". Judging by their responses, they are both talented physics students and I am grateful for their assistance. However, I do not completely understand the working given by "PPTR" for questions 2.3.2 b and 2.3.2 c and was just simply asking for a further explanation (as I believe most people do when they dont completely understand something).

Furthermore, I created this thread with the intention of receiving a little bit of help with the above questions. Nowhere is there any mention of "please abuse me for not being the best physics student there ever was" so may I suggest you stop being the smartass you are coming across as and actually help me with these questions if you have any relevant physics knowledge. If not, shut the fuck up and find something better to do.

Calm your tits.
There are assholes in every forum and you've experienced trolling on in here.

 

[Bored Of Fail 2]

Bored of Fail 2, why is it that we take down to be the positive direction? I havnt really heard of doing this although it did prove useful for this question. Also, I hope sometime in the near future you can "feel like" answering the remaining questions for me.

.

When doing a projectile question you should draw a picture of the situation , and usually, you take the inital direction of the projectile as the " positive y" direction , in the case of the first question I answered, the initial velocity of the ball was downwards, so I chose downwards as my "positive y" direction , and thus why I gave acceleration ( gravity ) a positive value , because it acts in the direction which I have taken to be positive.

However, in the third question it is different, the initial motion of the projectile is upwards, so I take that as my positive y direction, and since gravity acts down towards the earth ( ie in the nagetive y direction ) I gave it a negative value

always draw a picture and assign a positive y direction, and if acceleration acts in the same direction then give it a positive value , otherwise give it a negative value

 

[Bored Of Fail 2]

downwards is considered positive.

Substitute the values: 0 = u.y + (-9.8)(3.75); Therefore, u.y = 36.75 m/s

Substitute 36.75 into the eqn r = (u*t) + (1/2)*a*(t^2); that should give you the maximum height.

c) From parts a and b, you have calculated the horizontal and vertical components. Find the initial velocity and the angle using simple trignometry.

Hope that was helpful...


Need Help??? Check this out....
http://community.boredofstudies.org/showthread.php?t=254227

thats prob what is confusing the OP, should say downwards is negative, thus acceleration = -g = -9.8m/s^2

 

[LogE]

I doubt that OP understands even now after you have given ur explanation....

 

[clissold]

Calm your tits.
There are assholes in every forum and you've experienced trolling on in here.

Ohh, well spotted captain obvious..

If you didnt know, Im not oblivious to trolling. Its just that losers such as "LogE" need to be taught a lesson, and should find something better to do with their time.

Thankyou anyway.

When doing a projectile question you should draw a picture of the situation , and usually, you take the inital direction of the projectile as the " positive y" direction , in the case of the first question I answered, the initial velocity of the ball was downwards, so I chose downwards as my "positive y" direction , and thus why I gave acceleration ( gravity ) a positive value , because it acts in the direction which I have taken to be positive.

However, in the third question it is different, the initial motion of the projectile is upwards, so I take that as my positive y direction, and since gravity acts down towards the earth ( ie in the nagetive y direction ) I gave it a negative value

always draw a picture and assign a positive y direction, and if acceleration acts in the same direction then give it a positive value , otherwise give it a negative value

Thankyou, finally some genuine assistance

I have found that drawing a picture of the situation does help so thankyou for that suggestion. Also, your explanation of finding a "positive y" direction has been very helpful too.

If I may clarify, whichever direction the projectile initially moves in should be the direction I assign as "positive". For example, when a projectile is fired off a cliff, its initial direction is downwards and thus I should take down as being my "positive" direction. Just as if a projectile is fired at an angle above the horizontal, I should take up as being my "positive" direction as this is the direction in which the projectile initially moves.

 

[deterministic]

Just as if a projectile is fired at an angle above the horizontal, I should take up as being my "positive" direction as this is the direction in which the projectile initially moves.

Only consider vertical direction (up/down) as you will not deal with any horizontal forces in projectiles.

It actually does not matter which direction is positive, as you just label anything going opposite to that direction to be negative. But for convenience sakes, do as you have said.

 

[Bored Of Fail 2]

For example, when a projectile is fired off a cliff, its initial direction is downwards and thus I should take down as being my "positive" direction.

no, not always , it is just that in the case of the first question you had here, the projectile was fired "horizontally" from a cliff , it had no initial vertical velocity , so since it was not fired upwards it would tend to initially move downwards under the influence of gravity, thus why I chose downwards as the positive direction.

It is possible to have a question where lets say a projectile is fired from a cliff 100m high at a initial speed of 20m/s at an angle of 20 degrees to the horizontal, in this case there is a initial y velocity that is causing it to move up, so in this situation we would take positive y direction as upwards, and thus acceleration would be -g as it acts against the projectiles initial motion.

 

[clissold]

and btw OP, finish ur own HSC first before teaching anything to anyone

I havnt tried to explain anything to anyone, all I did was clarify in order to attain my better understanding of projectile motion

 

[clissold]

"Bored of Fail 2" ... THANKYOU

you have been of a great amount of help, the only questions I am unsure of now are 2.3.2 b and 2.3.2 c

P.S.. I sure hope you are not doing your HSC this year, hahahahaha

 

[Bored Of Fail 2]

"Bored of Fail 2" ... THANKYOU

you have been of a great amount of help, the only questions I am unsure of now are 2.3.2 b and 2.3.2 c

P.S.. I sure hope you are not doing your HSC this year, hahahahaha

nah i did it in 2009 , im onto 2nd year electrical engineering at unsw this year