# Some acceleration problem?

Is it -9.81m/s2?????

if so i need some explanations!

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!

Is it -9.81m/s2?????

if so i need some explanations!

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active| newest | oldestYou 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)

in both ways

^{-2}) you also have aerodynamic drag, which is a cube-law and dependent upon the size of the ball etc...L

There is only one force acting on the body.

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)

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.