# Physics question, really simple.

Discussion in 'Community Discussion' started by Jasonbot, Nov 14, 2007.

1. ### Jasonbot macrumors 68020

Joined:
Aug 15, 2006
Location:
The Rainbow Nation RSA
#1
I'm writing my end of year physics exam tomorrow and need help with this problem.

A ball is thrown vertically upwards into the air. It's flight lasts 2.0s. what was the maximum height that th ball reached?

It should=5m. But I just cant get that.

2. ### Much Ado macrumors 68000

Joined:
Sep 7, 2006
Location:
UK
#2
You need initial velocity. Use the magic of the Suvat equations.

3. ### LtRammstein macrumors 6502a

Joined:
Jun 20, 2006
Location:
Denver, CO
#3
Use your force and projectile motion equations!

Also, you need to make sure you have all the initial conditions.

2s = t
9.81 m/s^2 = g
? = vy0 (initial vertical velocity)

4. ### .JahJahwarrior. macrumors 6502

Joined:
Jan 1, 2007
#4
well, d=.5(Vi+Vf)*deltaT

Hm...you need to know the initial velocity to use that one, though.

I know there are tonnes of other equations, but I don't have my 3x5 card of the best equations with me any more. With that card, I could solve most any problem....

Joined:
Oct 11, 2004
Location:
On the Left Coast - Victoria BC Canada
#5
Nope. You don't need to know initial velocity. Assuming no resistance, the ball goes up for one second, reaches zero velocity at the top, and then down for one second. Therefore you can simply solve for the distance because you know the time (1 second) and the acceleration (gravity) - just like you would a dropped object.

6. ### Jasonbot thread starter macrumors 68020

Joined:
Aug 15, 2006
Location:
The Rainbow Nation RSA
#6
Thats teh problem. None of these are given but It's multiple choice and my options are 5,0,20 or 40. I keep getting 20m

Using taht logic the ball should accelerate I say after 1s it goes 10m up and after 2s it comes back to rest reaching a max of 10m.

7. ### Much Ado macrumors 68000

Joined:
Sep 7, 2006
Location:
UK
#7
Right, just worked this out:

v= u +at
0=u-10

u=10 (Obvious)

With that:

s=ut +0.5att
=10-10/2
=5

EDIT: CanadaRAM- I played it safe and took 'a' to be -10m/s/s

Joined:
Oct 11, 2004
Location:
On the Left Coast - Victoria BC Canada
#8
Reaching a max velocity of about 10 m/sec you mean. But it doesn't reach a height of 10 m, because the distance is the average of the velocities, times the time. Remember the problem isn't asking how far the ball travelled, but how high it reached. Therefore you only consider one leg of the journey

Vo = 0 (where it stops at the top of the arc)
Vf = Vo + A*T = 0 + (9.81 x 1) = 9.81 (the downward speed it's going as it returns to the thrower)
T= 1 sec

D= (Vo + Vf)/2 * T

D= (0 + 9.81 m/sec)/2 * 1 sec = 4.905 m

THe correct answer would be 4.9 m higher than the height from which it was thrown.

Frack - it's been 30 years since I did this @%\$* !! Don't make me think so hard !!

9. ### Jasonbot thread starter macrumors 68020

Joined:
Aug 15, 2006
Location:
The Rainbow Nation RSA
#9
Thanks guys! I can see a 100% coming nearer!!!

10. ### mgargan1 macrumors 65816

Joined:
Feb 22, 2003
Location:
Reston, VA
#10
Shouldn't it just be 9.8/2?

Acceleration of earth's gravity... 9.8ms it's going up for 1 second, and coming down for one second...

you only need to know how much it comes down, so, just divide by two...

11. ### Jasonbot thread starter macrumors 68020

Joined:
Aug 15, 2006
Location:
The Rainbow Nation RSA
#11
We use gravity as 10.

12. ### Jasonbot thread starter macrumors 68020

Joined:
Aug 15, 2006
Location:
The Rainbow Nation RSA
#12
Well as Murphy's Law states, anything that can g wrong will go wrong and unfortunately I never even received a question remotely like this one in my exam. But the rest of the exam was so easy, even though other people said that they were dying. yay