kinematics

Jiemanutz SG

The hot air balloon is now travelling vertically upward at a constant speed of 5 m/s. When it is 30m above the ground,a second package is released from the balloon.

How long does the second package take to reach the ground?

What is the speed of the package just b4 it hits the ground?

thanks

iamapebble SG

Taking upwards and right as positive, and since any object dropped from a moving object will have a initial velocity equals to the moving object,

We have:
s = ut + 1/2at^2
-30 = 5t + 1/2(-9.81)t^2
   .
   .
   .
t = 3.03s or t = -2.01s (rejected because time cannot be negative)

Next, speed just before the package hits the ground is actually the final velocity of the package, therefore we have:

a = (v-u)/t
-9.81 = (v - 5)/3.03s
v = -24.7ms^-1 (velocity is negative due as it is moving downwards)

However, since question requires speed, we can just leave as:

Speed of package just before it hits the ground = 24.7ms^-1.

Darkness_hacker99 SG

Speed of the package just before hit the ground (assume negligible air resistence):

Let v be the velocity, h be the height, g be gravitation constant

v = sqrt 2gh.

G acceleration. 9.8m/s (assuming). Meaning every second, it's adds 9.8m to the distant it dropped. ie. 9.8, 19.6, etc.

Time taken for the second object to fall down = 30/9.8m/s = 3.06 seconds.

P.S I long time didn't do physics le. dunno whether correct. Those who know, please correct me alright >.<" Thanks~

Jiemanutz SG

the answers are 3.03s and 24.m/s respectively

thanks all

Darkness_hacker99 SG

I learned something new today. :)

eagle SG

well done iamapebble

blueftw SG

how does the mass of a projectile affect it's range??

eagle SG

how does the mass of a projectile affect it's range??

it doesn't

the initial velocity affects it

16/f/lonely SG

it doesn't

the initial velocity affects it

it doesn't provided air resistence is ignored. Laughing

eagle SG

it doesn't provided air resistence is ignored. Laughing

yup

i forgot about air resistance, or rather, fluid resistance

Jiemanutz SG

how do we know when to write the correct sign convention for displacement and acceleration? wad is the use of a reference point?

thanks

16/f/lonely SG

how do we know when to write the correct sign convention for displacement and acceleration? wad is the use of a reference point?

thanks

There's no correct/wrong direction.

That's why there's a reference point. Laughing

Your answer can be negative/positive.

blueftw SG

so in conlclusion you all are saying that when there is air resistance, the mass of the projectile affects the range?

crimson soldier SG

so in conlclusion you all are saying that when there is air resistance, the mass of the projectile affects the range?

when there is air resistance, ma= mg-bv^2, solve the second order differential equation to obtain x as a function of t, and then differentiate it once again to get v!

....no wait this is O levels so leave out air resistance please : D

16/f/lonely SG

so in conlclusion you all are saying that when there is air resistance, the mass of the projectile affects the range?

Logically speaking.

Of course!

If you throw a ping pong ball and a fish ball, chances are the fish ball goes further.

Because the air particles have to do more work on the thing you throw if its heavier to slow it down the same amount as the lighter object because of momentum. F= ma, Momentum = mV.

The presence of air resistence complicates things further. You have to take surface area into consideration as well. It does not just stop at the area that is facing the direction you threw the object. Because of friction between the air particles and the object, obviously, the larger the surface area, the more friction and hence resistence. Not to mention, the less laminar the flow of the air around the object, the higher the resistence due to turbulence.

Ok, I shall stop here. Laughing

eagle SG

Logically speaking.

Of course!

If you throw a ping pong ball and a fish ball, chances are the fish ball goes further.

Because the air particles have to do more work on the thing you throw if its heavier to slow it down the same amount as the lighter object because of momentum. F= ma, Momentum = mV.

Ok, I shall stop here. Laughing

err... the first part is about giving it initial velocity... not really part of projectile motion, but more of on Newton's 2nd law