physics help

FrostedNipple · 11 Apr 2007 at 15:51

errrr golf ball being hit from the ground at an angle of 36degrees -- starting velocity is 42m/s -- if the two points are a and b its takes 5 seconds to get from a to b -- ignore air resistance --

i worked out that horizontal component of the velocity is 34m/s and its vertical component as 24.67

i have to work out the max height reached

i used s=ut+1/2at^2 and got 24.67*5 + 1/2 * 9.81 * 25

and i got 3171.425m now i didn't know golf balls went 3km in the sky, so I'm guessing I've either used the wrong equation or done something horribly wrong, please help!!!

Psyk · 11 Apr 2007 at 15:53

mavity acts in the opposite way to the ball's movement.

Visage · 11 Apr 2007 at 15:54

FrostedNipple said:
errrr golf ball being hit from the ground at an angle of 36degrees -- starting velocity is 42m/s -- if the two points are a and b its takes 5 seconds to get from a to b -- ignore air resistance --

i worked out that horizontal component of the velocity is 34m/s and its vertical component as 24.67

i have to work out the max height reached

i used s=ut+1/2at^2 and got 24.67*5 + 1/2 * 9.81 * 25

and i got 3171.425m now i didn't know golf balls went 3km in the sky, so I'm guessing I've either used the wrong equation or done something horribly wrong, please help!!!

The time at which the ball attains maximum height is *half* the total flight time.....also a is not +9.8, its -9.8, since mavity acts *downwards*.

Sleepy · 11 Apr 2007 at 16:30

Visage said:
The time at which the ball attains maximum height is *half* the total flight time.....also a is not +9.8, its -9.8, since mavity acts *downwards*.

t is not required to work out max h.

Use v^2=u^2 - 2aS

at max h, vertical velocity is 0, by definition. so u = 42cos54, v = 0, a = g, solve for s to obtain max height.

FrostedNipple · 11 Apr 2007 at 16:38

thanks everyone, lol, i knew it would only be stupid mistakes

Visage · 11 Apr 2007 at 16:39

Sleepy said:
t is not required to work out max h.

Use v^2=u^2 - 2aS

at max h, vertical velocity is 0, by definition. so u = 42cos54, v = 0, a = g, solve for s to obtain max height.

Very true. I was pointing out the flaw in *his* method though. To suggest a better method is simply doing his homework for him.....

Ricochet J · 11 Apr 2007 at 17:08

I would've have done a SUVAT table for verticle and horizontal componants and seen what one I needed (vertical height).

Arsey · 11 Apr 2007 at 17:50

I hate it how you have to ignore air resistance, so the answer you get is totally wrong anyway :mad:

ArmyofHarmony · 11 Apr 2007 at 17:54

Visage said:
Very true. I was pointing out the flaw in *his* method though. To suggest a better method is simply doing his homework for him.....

yeah but if he's revising and not doing homework, i'd just want to know what to do (I assume hes revising, since its easter... and I assume thats AS level PHysics module 1, an exam prob coming up)

daz · 11 Apr 2007 at 17:54

Draw a speed time graph for the vertical velocity component. The gradient of the graph will be 9.8m/s/s and there will be a turning point when the velocity is zero. At this point, look at the value of t. Then multiply this value of t by the horizontal velocity.

I find drawing the speed time graph much easier than remember arbitary formulae. You can see everything very clearly, and derive the formulae from there.

Ricochet J · 11 Apr 2007 at 17:55

Arsey said:
I hate it how you have to ignore air resistance, so the answer you get is totally wrong anyway

THen you might as well include the small change in mass as velocity increases to reach max height.

daz · 11 Apr 2007 at 17:55

Arsey said:
I hate it how you have to ignore air resistance, so the answer you get is totally wrong anyway

How would you calculate the air resistance force on a dimply golf ball?

Sleepy · 11 Apr 2007 at 18:05

daz said:
How would you calculate the air resistance force on a dimply golf ball?

With great difficulty ...

Rebelius · 11 Apr 2007 at 18:19

daz said:
Draw a speed time graph for the vertical velocity component. The gradient of the graph will be 9.8m/s/s and there will be a turning point when the velocity is zero. At this point, look at the value of t. Then multiply this value of t by the horizontal velocity.

I find drawing the speed time graph much easier than remember arbitary formulae. You can see everything very clearly, and derive the formulae from there.

the max height is going to be half the total flight time though - it's a parabolic flight since he's ignoring air resistance and (assuming) it starts and ends at the same height.

v^2=u^2+2as is the better way to do it though, because then you don't need to know anything about where it lands.

qwerty · 11 Apr 2007 at 18:24

daz said:
How would you calculate the air resistance force on a dimply golf ball?

Actually it's fairly simple. I answer this is a golfer rather than a physicist:

Air resistance can only act on the surface area of the ball which is in contact with the air. For some reason, the turbulence caused by the dimples effectively means that they do not come into contact with the air so the only part of the ball subject to air resistance is the space between the dimples.
Originally golf balls were purely spherical, but golfers realised that they flew much further after they had been bashed up and scuffed, and eventually perfected this idea by making balls dimpled.

Sleepy · 11 Apr 2007 at 18:24

daz said:
Draw a speed time graph for the vertical velocity component. The gradient of the graph will be 9.8m/s/s and there will be a turning point when the velocity is zero. At this point, look at the value of t. Then multiply this value of t by the horizontal velocity.

That doesn't produce max height.

I find drawing the speed time graph much easier than remember arbitary formulae. You can see everything very clearly, and derive the formulae from there.

Every method requires you remembering something. You can simply quote the various SUVAT equations or derive them from first principles. In this case starting from:

∆(KE + PE)=0

you can derive v=u+2aS

daz · 11 Apr 2007 at 18:24

Rebelius said:
the max height is going to be half the total flight time though - it's a parabolic flight since he's ignoring air resistance and (assuming) it starts and ends at the same height.

It's a parabolic flight, but the vertical velocity versus time graph can drawn as a straight line. And since you have symmetry, you only need to draw half the graph.

Easy.

daz · 11 Apr 2007 at 18:26

Sleepy said:
That doesn't produce max height.Every method requires you remembering something.

It's a diagonal straight line with a gradient of the acceleration due to mavity from the left hand axis at max velocity, to the other axis, time when v=0.

The area under the graph is the distance travelled vertically.

I've drawn a picture for you.

So the gradient is 9.8 m/s/s.

24.67m/s divided by 9.8m/s/s = 2.52 seconds (this is when v = 0, i.e. halfway through the parabolic path travelled by the ball)

To find the area of the graph, and thus the distance travelled you do
1/2*(24.67m/s * 2.52 seconds)

= 31 Metres as the maximum height.

That 2.52 seconds is also the time until the ball reaches its apex. The total flight time will therefore be twice this. You can multiply this by the horizontal velocity to find the horizontal distance travelled (since the horizontal velocity is assumed constant).

To the OP: I hope the explanation and graph has helped you grasp a better understanding of the process involved, rather than simply memorising some formulae.

Sleepy · 11 Apr 2007 at 18:32

daz said:
The area under the graph is the distance travelled vertically.

[homer]DOH!!![/homer]

Rebelius · 11 Apr 2007 at 18:34

s = 1/2 * a * t^2, no?

my teachers been lying to me?