Some acceleration problem?

Lets say that a  ball of unknown mass is projected with and unknown force vertically upwards with a velocity of 40m/s. What's its deceleration???
Is it -9.81m/s2?????

if so i need some explanations!

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frollard5 years ago
***ALL OF THE FOLLOWING IS ASSUMING A SPHERICAL COW WITH NO AIR RESISTANCE ON A FLAT PLANET***

You can only express the force as a function of the mass if you don't know the mass.
(force = mass * acceleration), in the units of kg*m/s^2
and acceleration = force / mass

There is no such thing as deceleration, only negative acceleration for the purposes of the formulas.

So, its got an upward (presuming 0º from normal, vertical) at +40 m/s.

It will accelerate downward at +9.81m/s, but we consider up to be positive, so 'upward' at -9.81m/s^2.

One neat thing is it doesn't matter how much mass it has, because a force will be proportional to said mass.
You can use lots of the uniformly accelerated motion formulas
like (they all derrive to one another so you can interchange lots of them)
acceleration = (final velocity - initial velocity) / time
a = (Vf - Vi)/t
t = (Vf - Vi)/a
t = (-40m/s - 40m/s)/-9.81m/s^2
t = -80m/s / -9.81m/s^2
t = 8.1549 seconds
2 significant figures.
t = 8.2 seconds if hang time between launch and impact
or 4.1 seconds to apogee.

Now with acceleration and time, or the longer formula of the velocities and time, we can figure out the height (distance)
Vf^2 = Vi^2 + 2ad
0^2 = (40m/s)^2 + 2(9.81m/s^2)d
(-40m/s)^2 = 2(9.81m/s^2)d
d = (-40m/s)^2 / (19.62m/s^2)
d = 1600m^2/s^2 / 19.62m/s^2
d = 81.5494m
2 significant figures
d = 82 meters.

The force acting on it (in newtons) can only be calculated based on its mass.
force = mass * acceleration
a newton is a KG*m/s^2
For example
if the object were 100kg
assuming its in motion already (we don't have to impulse it to speed)
the force of gravity on the object would be
f = ma
f = 100kg * 9.81m/s^2
f = 981 newtons.
At maximum speed (40m/s) it would have kinetic energy (Ek) in joules of:
Ek = 1/2 mv^2
Ek = 1/2 100kg*(40m/s)^2)
Ek = 1/2 * 100*1600
Ek = 80,000 Joules = 80kJ

At maximum height, zero velocity it would have the same potential energy (it is conserved)
Ek(launch) = Ep(apogee) = 80kJ.

At maximum speed, it would have a momentum (p) (Newton-seconds)
p = mass * velocity
p = 100kg * 40m/s
p = 4000 kg*m/s or 4.0 kNs)
Kiteman5 years ago
In the real world, or in the world of Physics homework?
ARJOON (author)  Kiteman5 years ago
LOL!!! i think you're a teacher,


in both ways
lemonie5 years ago
Apart from gravity (-9.81ms-2) you also have aerodynamic drag, which is a cube-law and dependent upon the size of the ball etc...

L
What forces act on the ball ? Since you don't qualify the question with any requirements to consider air friction.....

There is only one force acting on the body.
ARJOON (author)  steveastrouk5 years ago
OK, now lets say we have a the mass X, and force Y,

and the ball is projected from rest with force Y applied
There must be a time where the acceleration is positive and then becomes negative (-9.81)
It only accelerates when the Force Y is applied.
Ignoring air resistance, and assuming it stays close to the surface of the earth, and is not travelling at a significant fraction if the speed of light, It will always accelerate at 9.81m/s/s towards the earth.

In your example, where up is positive, and down is negative, the ball will always accelerate at -9.8m/s/s. Deceleration is the rate of decrease of velocity, so as an object slows, its deceleration is positive.