# kinematics question

• Oct 13th 2006, 03:20 PM
scorpion007
kinematics question
A balloon is rising vertically at a constant speed of 21 m/s. A stone is dropped from the balloon when it is h metres above the ground. The stone strikes the ground 10 seconds later. Assuming that air resistance is negligeble, what is h?

this probably isnt a very complex question, im probably just forgetting something trivial...
• Oct 13th 2006, 06:35 PM
Quick
Quote:

Originally Posted by scorpion007
A balloon is rising vertically at a constant speed of 21 m/s. A stone is dropped from the balloon when it is h metres above the ground. The stone strikes the ground 10 seconds later. Assuming that air resistance is negligeble, what is h?

this probably isnt a very complex question, im probably just forgetting something trivial...

note:upwards is positive

(starting height) = (ending height) - (starting velocity)(time) + (1/2)(acceleration)(time)^2

thus: h = 0 - (21)(10) - (1/2)(-9.8)(10)^2

then: h = (1/2)(9.8)(100) - 210

multiply: h = 490 - 210

finally: h = 280

Although in your work you should label each number (with meters, seconds etc..)
• Oct 13th 2006, 11:48 PM
scorpion007
thank you very much, i wasnt aware of this forumula, as my math text book doesn't cover it. It appears to only have the 5 classic ones.
EDIT:
although, why did you go from
(starting velocity)(time) + (1/2)(acceleration)(time)^2
to:
(21)(10) - (1/2)(-9.8)(10)^2

i.e. why did you change the + to a - ?

Also, could i have solved it using the classic formula's? such as: s = ut + 1/2at^2 ?
• Oct 14th 2006, 03:22 AM
Quick
Quote:

Originally Posted by scorpion007
thank you very much, i wasnt aware of this forumula, as my math text book doesn't cover it. It appears to only have the 5 classic ones.
EDIT:
although, why did you go from
(starting velocity)(time) + (1/2)(acceleration)(time)^2
to:
(21)(10) - (1/2)(-9.8)(10)^2

i.e. why did you change the + to a - ?

I defined upwards to be positive, the rock started out going 21 m/s up, thus it was positive

However, gravity is going downward, thus the 9.8 is negative.

Quote:

Also, could i have solved it using the classic formula's? such as: s = ut + 1/2at^2 ?
probably, but you'll have to tell me what s and u stand for...
• Oct 14th 2006, 06:23 AM
topsquark
Quote:

Originally Posted by scorpion007
A balloon is rising vertically at a constant speed of 21 m/s. A stone is dropped from the balloon when it is h metres above the ground. The stone strikes the ground 10 seconds later. Assuming that air resistance is negligeble, what is h?

this probably isnt a very complex question, im probably just forgetting something trivial...

What Quick did was okay, but he got some postives and negatives screwed up. I'm not razzing him for it, it's a common mistake.

For starters, Quick WAS using your formula. Apparently he didn't know the variables so he used words instead.
(Edit: I looked at Quick's work again and, for some reason, the equation he was using has an extra negative sign all throughout it. Mathematically it is the same as the displacement equation used here, but in terms of the Physics I can't make any sense out of it.)

Let me run you through this the way I would set it up.

Picture: We have a balloon rising at a velocity of 21 m/s upward. A rock is dropped from rest with respect to the balloon, so the initial velocity of the rock is the same as the balloon's: 21 m/s upward. The rock hits the ground at time t = 10 s.

Coordinate system: I'm going to choose +y upward and I'm going to choose the origin (y = 0 m) as being on the ground.

I'm going to use my variation on s = ut + (1/2)at^2. s is a displacement and as we are looking for distances I'm going to write s = y - y0, where y(t) is the height of the rock at a given time. Thus y = y(t) and y0 = y(t = 0). u represents the initial velocity of the object. Usually that is written as u = v0. So the equation becomes:
y - y0 = v0*t + (1/2)at^2
I don't know how your textbook presents this equation (presumably s = ut + (1/2)at^2) but this form of the equation is rather more common.

So. We know:
t = 10 s
y = 0 m (the rock is on the ground at t = 10 s)
y0 = ? (This is what we are looking for)
v0 = 21 m/s
a = -g = -9.8 m/s^2 (g is NEVER negative!)

0 - y0 = 21*10 + (1/2)(-9.8)(10)^2

-y0 = -280

y0 = 280 m.

-Dan
• Oct 14th 2006, 08:17 AM
Quick
Quote:

Originally Posted by topsquark
What Quick did was okay, but he got some postives and negatives screwed up. I'm not razzing him for it, it's a common mistake.

For starters, Quick WAS using your formula. Apparently he didn't know the variables so he used words instead.
(Edit: I looked at Quick's work again and, for some reason, the equation he was using has an extra negative sign all throughout it. Mathematically it is the same as the displacement equation used here, but in terms of the Physics I can't make any sense out of it.)

Quote:

Originally Posted by topsquark
If we assume that the acceleration is constant the equation of motions are:
y = y_0 + v_0t + (1/2)at^2

Quote:

Originally Posted by topsquark
y and y0 are the final and initial positions (y = y(t), y0 = y(0).)

I was solving for y_0,

so -y_0 = v_0t + (1/2)at^2 - y

multiply by -1 to get: y_0 = y - v_0t - (1/2)at^2
• Oct 14th 2006, 09:12 AM
topsquark
Quote:

Originally Posted by Quick
I was solving for y_0,

so -y_0 = v_0t + (1/2)at^2 - y

multiply by -1 to get: y_0 = y - v_0t - (1/2)at^2

Ah, I thought that was your starting equation. That makes sense then. :)

-Dan
• Oct 14th 2006, 04:24 PM
scorpion007
i see, thanks for all the help guys