Physics question, really simple.

[Jasonbot]

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.

 

[Much Ado]

You need initial velocity. Use the magic of the Suvat equations.

 

[LtRammstein]

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)

 

[.JahJahwarrior.]

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....🙁

 

[CanadaRAM]

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.

 

[Jasonbot]

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

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.

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.

 

[Much Ado]

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 🙂

 

[CanadaRAM]

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.

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 !! 🙂

 

[Jasonbot]

Thanks guys! I can see a 100% coming nearer!!!😀

 

[mgargan1]

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...

 

[Jasonbot]

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...

We use gravity as 10.

 

[Jasonbot]

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😀