• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Physics virgin need help on freely falling objects!

  • Thread starter noodle21
  • Start date

noodle21

You drop a ball from a window on an upper floor of a building. it strikes the ground with velocity v. You now repeat the drop, but have a friend down on the street who throws another ball upward at velocity v. Your friend throws the ball upward at exactly the same time that you drop yours from the window. At some location, the balls pass each other. Is this location at the halfway point between window and ground, above this point, or below this point?
 

krab

Science Advisor
896
1
1. You are in the wrong forum. This one is Quantum Mechanics. There is a forum for help with college physics.

2. Anyway, your friend's ball starts upward with a velocity v, but your ball starts with velocity = 0. So whose ball gets to the halfway point first? Think about it.
 

HallsofIvy

Science Advisor
Homework Helper
41,709
876
Originally posted by noodle21
You drop a ball from a window on an upper floor of a building. it strikes the ground with velocity v. You now repeat the drop, but have a friend down on the street who throws another ball upward at velocity v. Your friend throws the ball upward at exactly the same time that you drop yours from the window. At some location, the balls pass each other. Is this location at the halfway point between window and ground, above this point, or below this point?
Originally posted by krab
1. You are in the wrong forum. This one is Quantum Mechanics. There is a forum for help with college physics.

2. Anyway, your friend's ball starts upward with a velocity v, but your ball starts with velocity = 0. So whose ball gets to the halfway point first? Think about it.
Well, apparently it got moved! In your second comment, you seem to be implying that since the ball thrown up starts with speed v while the ball dropped starts with speed 0, the ball thrown will "move faster". Don't you think acceleration (and deceleration) will have something to do with it?

Suppose the window is h meters above the ground. Both balls have an acceleration of -g. The ball dropped will have speed (negative so downward) of v1(t)= -gt and the ball thrown upward will have speed (positive so upward) v2(t)= v- gt.

The height, at time t, of the dropped ball is
h1(t)= -(g/2)t2+ h while the height, at time t, of the thrown ball is h2(t)= vt- (g/2)t2.

However, v is not just some arbitrary speed. It is the speed the dropped ball has when it hits the ground: The dropped ball hits the ground when h1(t)= (-g/2)t2+ h= 0 or when
t= [sqrt](2h/g). At that time, v= -g([sqrt](2h/g)= -[sqrt](2hg).
Of course, the speed of the ball thrown up is [sqrt](2hg).

They will pass when they both have the same height at the same time:
-(g/2)t2+ h= [sqrt](2hg)t- (g/2)t2.
By golly, the accelerations cancel out! This reduces to
[sqrt](2hg)t= h so t= h/[sqrt](2hg)= [sqrt](h/2g).

At that time, h1= (-g/2)(h/2g)+ h= -h/4+ h= (3/4)h.
Of course, h2= [sqrt](2hg)[sqrt](h/2g)-(g/2)(h/2g)
= h- h/4= (3/4)h.

Well, I'll be! krab was right! The ball thrown upward, because it had the greater initial speed (and we DIDN'T have to take acceleration into account- it canceled out) goes farther. The two balls pass 3/4 of the way up to the window.
 

Related Threads for: Physics virgin need help on freely falling objects!

  • Posted
Replies
1
Views
988
Replies
1
Views
2K
  • Posted
Replies
1
Views
2K
  • Posted
Replies
1
Views
1K
  • Posted
Replies
5
Views
2K
Replies
4
Views
8K
Replies
6
Views
15K
Replies
3
Views
681

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top