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YEs, I thought of that... but don't know the mass of the ball?

Hi all, I know this may see basic but how would I go about calcualting Impact velovity? The only information I have is: Height dropped, height bounced and from here I can calculate the CofR... But how would I go about finding impact velocity

Calculate the pull on the high bar of a gymnast performing a giant swing. His angular velocity is 2 rads, his mass 80 kg and effector radius of rotation (distance from bar to CofG) is 1 meter Hi... I am a bit confused... "pull on the bar" would this be the work done on the bar to the person...

So if he's moving at a constant velocity --- he's not accerating therefore threre is no force affecting his velocity?

An athlete (mass of 50 kg) is moving with a constant velocity of 5ms. Determine the resultant force acting on him. If it's a constant velocity he is not acclerating? Therefore a = 0 m/s/s; F=ma F=50 X 0 = 0N? But there has to be a force dosent there? Would the force be equal to the mass...

yes... get it now... fwwwww lol... thanks for all you help!

(Q) What torque is applied to a steering wheel of an automobile when the driver applies a force of 50 Newtons tangent to the circumference of the wheel? Diameter of the steering wheel is 40 cm. F = 50 N D (torque arm) = 0.40m T= Fd = 20 N... but its wrong... any ideas...

Yes but it says what is her maximum velocity: So v=sqrt 2gh @ 1m = 4.43 m/s but v=sqrt 2gh @ 2m = 6.26 m/s ... (so this does not make sense to me...)

So then this indicates that the max velcoity is at the higest point? I though max velcoity is at the lowest point??... but her velocity @ 1m using v= sqrt 2gh also = 4.43 m/s... so is the velcoity then constant?

okay well.. the answer is 4.43 m/s... so v=sqrt (2 x 9.81 x 2) = 4.43m/s... aff yeas on reading the question it say's her velocity at highest point... Thanks for all your help... I undersatnd how to getthe answer ..but confused how it can be solved without a mass...

rayquesto ... but this would make here velocity at the highest point faster than the lowest point... My undersatdning is velocity is at it/s max when d=1m...

Okay, if PE = 9.81 J @ 2m above the ground... and energy is conserved, @1m above the ground all energy will become KE?; KE=9.81J? if I plug that into KE=1/2mv(sqr) I still can't comput an answer... I don't have mass (sorry... this is fustraing the ! out of me lol)... I understand...

Hi all thanks for all you help... PeterO... the equations 1/2mv(sqr)= 0 = KE @ 2m? and mgh=0 = PE @ 1m? so.. PE + KE = C would be: mgh + 1/2mv(sqr) = c? im really confused... I don't think the question is intended to be this complicated... I think its trying to gte me to think about PE...

All I can think of is the realtionship between energy conservation... So at 2m (max height ) Her PE = max and at 1m (min) her KE=max where her velocity is max and her PE = min/0j) If I use PE=mgh at 1 m; 0=m x 9.81 x 1 = 9.81 kg... Then KE at 2m = 1/2mv(sqr) = 0=4.91v(sqr) = 2.2m/s...

The only issue i have using an energy formula is I don't have the mass? othwise I would use: PE + KE = C wt x h x 1/2 mv(sqr) = c then I would sub in C into: wt x h x 1/2 mv(sqr) = c, to find V...

okay, then the other hting I can think of is using 2piR 2 x pi x 10m = 62.83m ? Heading in the right direction?

So if her maximum velocity is at d-1m... Then?? I can only think of: acc due to gravity = 9.81 ms/sqr im stuck with how to approach the next step...

(Q) A girl is swinging back and forth on a swing suspended from a rope 10m long. At the lowest point of her swing she is 1m from the ground, and at the highest point she is 2m above the ground. What will be her maximum velocity? Would her maximum velocity be at the lowest point d = 1m above...

Ohhh thankyou... lol... (always read the question twice and double check you understand it lol)...

A 20 kg mass is sitting on the horizontal surface of a table top. Calculate the following Neglecting friction, calculate the force required to give the mass a velocity of 2ms. So far all I can think of is... F=ma m=20kg a=? a= Vf-Vi / t ... I am stuck ... all I can think of is F =...

(Q) If a diver (standing on a 15 meter high diving board) dives horizontally from the board, and mass is 50 kg. What is his vertical velocity 10 meters from the board ? PE @ 15m = mgh = 7358j (max) PE @ 10m = mgh = 4905j KE @ 10m = 7358 - 4905 = 2453j KE = 1/2 mv(sqr) 2453= 0.5 x...

Hi.. thanks what you have said makes sense... But is I use: wf=wi + at wf= 0 + 10 X 1.5 wf = 5 rad/s (sqR) The linear momentum = mv = 0.73 x 5 = 3.65 kg. m/s But the answer is 3.9 kg. m/s, I know I am only .25 off but is the book wrong or my calculations wrong ??

(Q) a 108 cm, 0.73kg golf club is swung for 0.5s with a constant acceleartion of 10 rad/s (squ). What os the linear momentum of the club head when it impacts the ball? Known: H= 108cm = 1.8m m = 0.73 kg =7.16N t = 0.5s \alpha= 10 rad/s (squ) I know momentum = mass x velcoity So...

A 35N hand and forearm are held at a 45 degree angle to the vertically oriented humerus. The centre of gravity of the forearm and hand is located at a distance of 15cm from the joint centre at the elbow, and the elbow flexor muscle attach at an average distance of 3cm from the joint centre. How...

Hi, I have the following q's. I can get to through most of it but I seem to get confused with computing the last part and get the wrong answer... can anyone help? Q. The knee extensors insert on the tibia at an angle of 30degree angle at a distance of 3cm from the axis of rotation at the...

A swimmer crossing a river proceeds at an absolute speed of 2.5 m/s on a course oriented at a 45degree angle to the 1 m/s current. Given that the absolute velocity of the swimmer is equal to the vector sum of the velocity of the current and the velocity of the swimmer with respect to the current...

Hi, I am just wondering if someone can help explain a concept of CG: I understand that in a homogeneous object the CG is in the geometric centre In a object without a constant mass, the CG will be towards the end with the greater mass But it say in my textbook that: "It is also possible...

yes but avg velocity = change n dispalcement over chnage in time... so his displacemnt = 0 ... but his chnage in time?

yes but avg velocity = change n dispalcement over chnage in time... so his displacemnt = 0 ... but his chnage in time?

Hi Tomer, thanks and yes I really would love to understand it for my exam.. Okay going back to cord. So Vx=horizontal velocity (which is constant ) = 20 m/s So Vy=vertical velocity (which changes by 9.8m/s) = 10m/s initial which is my Viy? Sorry i think I am just not used to your...

Hi Tomer, thanks for your help... the answer is 3.24s, But I just need to know how to get there... I have seen a few examples with the method you use which is confusing...I think in our Lab class it was a section of the range formula=v^2Sin\varthetaCos\vartheta +...

I agree... but apprently the answer is 0m/s...

[b]1. A handball is projected at an angle of 30 degrees with the horizontal from a 20 meter tower with an initial vertical velocity of 10 ms and an initial horizontal velocity of 20 ms. Find the total time of flight. Hi, I understand the concepts of projectiles when the landing height is...

sorry, yes that is the questions I understood it as each 400m lap he psoted those times... therefore will I need to average out the 4 times and work out the average velocity over a 1500m race?? So confused...

Homework Statement What is the average velocity of a runner whose 400 meter lap times are 60, 75, 65 and 60 seconds over a 1500 meter race? Homework Equations The Attempt at a Solution Is I use v=d/t (would the displacement be 0m beacuse he completed 1 lap? Im a bit...

Homework Statement A handball is projected at an angle of 30 degrees with the horizontal from a 20 meter tower with an initial vertical velocity of 10 ms and an initial horizontal velocity of 20 ms. Find the total time of flight. Homework Equations The Attempt at a Solution I...

"A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the angle of take off?" Okay, I did a skectch to work out basic trig, with 8m/s as my H and 0.5m as my 0 using sinϑ=o/h, =0.5/8 = 3 but it should be 21.6 ... do you know where I went...

Hi again... the second part to the question "A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the angle of take off?" Okay, I did a skectch to work out basic trig, with 8m/s as my H and 0.5m as my 0 using sinϑ=o/h, =0.5/8 = 3 but it...

would it just be as siple as 10m height @ 5m/s = 2s?

[b]1. A springboard diver performs a dive from a 10 metre tower and jumps with an initial vertical velocity of 5 ms and horizontal velocity of 2 ms. Determine the time she has in the air to perform her dive? The Attempt at a Solution From what I know 0=Vi + at (will give me 1/2 the...

oh thanks, so if I use 0=Vi^2 + 2ad 0=Vi^2 + (2 x 9.81 X .05) Vi^2=9.81 v=3.13 Would I be correct??

Hi, yea the exact words are "A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the upward velocity? " I assumed because its asking for upward velocity = Vertical component which is Vsin(theata) so Theta is the unknown? The next questions...

I know... I got 5.10m, whihc would give 7.1m... I think my teacher sometimes uses 10 or 9.81 depends what kind of day it is ;-), lol

Can anyone else help... I still can't get 0.14m? :-(

Homework Statement A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the upward velocity? Homework Equations The Attempt at a Solution Vv=Vsin\vartheta So \vartheta is the unkown Okay, I did a skectch to work out...

Homework Statement A baseball is thrown with an initial velocity of 20 ms at an angle of 30 degrees with the horizontal. How high did it go if it was released and caught 2 meters from the ground? Homework Equations The Attempt at a Solution 0=Vi^2 + 2ad 0=(20Sin30)2 + 2 X...

thanks for your help, that's what i was thinking would it then mean that the Vv is 5Cos20 instead of 5Sin20? Is that what you are suggesting?, also the answer is 0.14m...

Homework Statement A jumper takes off with a velocity of 5 ms at an angle of 20 degrees to the horizontal. How high does he raise his center of gravity? Homework Equations The Attempt at a Solution I know that Vh= 5scos20 = 1.7 m/s and Vv=5Csin20=4.70 m/s using 0=Vi^2 +...

Homework Statement A runner completes 6 1/2 laps around a 400m track during 12min reun test. What is the runners displacement at the end of the 12 min? Homework Equations The Attempt at a Solution Well the distance covered is 2.6km or 2600m in 720s, avg speed = 3.61m/s So I...

... oh dear... it is 8.3m/s ... no wonder I have spent 1/2 the day try to work it out ... lol... told you I would kickmyself... instead I think i need paperbag lol...