How do you calculate deceleration on vertically fired object?

Question

Taking out complex factors such as elevation, wind, etc. how do you figure out how much altitude an object gets and how fast does it decelerate?

I shoot a ball weighing 1kg (2.205 lbs) at 100 meters per second (328.1 fps)straight into the air. How far up should I expect it to go assuming no other factors hinder it?

Hope that's enough info to get the idea of what I'm trying to ask. Also, I'm not a physics major so please don't get too technical.

Answer 1

Hmmm...

Answer 2

this may well be a classic question that would trivially be spoke returned in case you assume that no capacity is lost. Kinetic capacity = a million/2 mass x speed skill capacity = mass x gravitational acceleration x substitute in precise particularly paintings out the precise the area the flexibility capacity won = kinetic capacity lost. be conscious that mass will cancel out indoors the real international you may would desire to difficulty approximately air friction yet i'm assuming which you're indoors the wonderfull international of element lots in a vacuum!

Answer 3

If shot straight up with no friction of any kind, you want to find the position of the ball when it stops going up, or when it's velocity drops to zero.

Since:
Velocity = InitialVelocity + Acceleration(Time)
Solve by substitution:
0 = 100 m/s + (-9.81m/s/s)(Time)
Time = 100/9.81 s = 10.19367992 seconds

Position is determined by:
Altitude = InitialAltitude + InitialVelocity(Time) + (Acceleration/2) ((Time)^2)
OR:
Altitude = 0 +100m/s(10.19367992 s) +((-9.81m/s/s )/2)((10.19367992) ^2)
SO:
Altitude = 509.6839959 meters

Answer 4

By conservation of energy: Ek = Ep

That is to say all of the mass' kinetic energy will eventually be transferred into potential energy.

1/2 m v^2 = mgh "m's" cancel

so the height, h, = v^2/2g

= 100^2/2x9.8

= 510.2m


Happy shooting!

Answer 5

This is a kinematics problem in Physics. Therefore, use the formula:

(Vf)^2 = (Vi)^2 + 2ad

your Vi (initial velocity) = 100m/s
Vf (final velocity) = 0
a is acceleration = -9.8 m/s^2

Put the numbers into the formula
0 = (100)^2 + 2 (-9.8)(d)
d = 510.20m

Hope this helps!

Answer 6

Everything that is on earth is subject to gravity, which creates an acceleration of -9.8 m/s/s on 'free fall' objects (objects that have no other forces acting on them, as in the situation you refer to).

Answer 7

use conservation of energy: KE=PE
1/2 m v^2 = m g h

h = v^2 / 2g (g=9.8 m/s^2)