# Projectile Motion

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

Two identical balls, X and Y, are at the same height and a horizontal distance of 25 cm apart.

X is projected horizontally with a velocity of 0.10 m s–1 towards Y at the same time that Y is released from rest. Both X and Y move freely in the absence of air resistance.

What is the distance between the balls 1.0 s later?

Can someone please help me with this question

Answer is 0.15m

X is projected horizontally with a velocity of 0.10 m s–1 towards Y at the same time that Y is released from rest. Both X and Y move freely in the absence of air resistance.

What is the distance between the balls 1.0 s later?

Can someone please help me with this question

Answer is 0.15m

0

reply

Report

#2

The vertical component of the motions of both balls will be the same, i.e. they will both be at the same height as each other at all times as they are both affected vertically by the same gravitational force. X will be moving horizontally with a constant velocity of 0.10m/s. So after 1 second how far has X travelled horizontally? Therefore how close are the two balls?

0

reply

Report

#3

Hm so i have 0.15m however I got it as a negative so I'm just trying to figure out why. So whenever you get these questions always do components and then resolve both vertically and horizontally for each particle. In this case, for particle Y, you can only resolve vertically as it is not projected horizontally.

(Original post by

Two identical balls, X and Y, are at the same height and a horizontal distance of 25 cm apart.

X is projected horizontally with a velocity of 0.10 m s–1 towards Y at the same time that Y is released from rest. Both X and Y move freely in the absence of air resistance.

What is the distance between the balls 1.0 s later?

Can someone please help me with this question

Answer is 0.15m

**Swb123**)Two identical balls, X and Y, are at the same height and a horizontal distance of 25 cm apart.

X is projected horizontally with a velocity of 0.10 m s–1 towards Y at the same time that Y is released from rest. Both X and Y move freely in the absence of air resistance.

What is the distance between the balls 1.0 s later?

Can someone please help me with this question

Answer is 0.15m

0

reply

Report

#4

X is projected towards Y. Its horizontal velocity is 0.1m/s and so the distance it travels is vt = 0.1*1=0.1m = 10cm. The original distance between them was 25cm and so the distance between them after 1s is 25 - 10 = 15cm.

Yes, for projectiles, you normally consider the horizontal motion separately from the vertical motion as the horizontal component is constant velocity and the vertical motion is constant acceleration due to gravity.

Yes, for projectiles, you normally consider the horizontal motion separately from the vertical motion as the horizontal component is constant velocity and the vertical motion is constant acceleration due to gravity.

2

reply

Report

#5

Ah I drew my diagram wrong hence i got a negative answer. I just assumed that X would be in front of Y which is obviously wrong.

(Original post by

X is projected towards Y. Its horizontal velocity is 0.1m/s and so the distance it travels is vt = 0.1*1=0.1m = 10cm. The original distance between them was 25cm and so the distance between them after 1s is 25 - 10 = 15cm.

Yes, for projectiles, you normally consider the horizontal motion separately from the vertical motion as the horizontal component is constant velocity and the vertical motion is constant acceleration due to gravity.

**Teenie2**)X is projected towards Y. Its horizontal velocity is 0.1m/s and so the distance it travels is vt = 0.1*1=0.1m = 10cm. The original distance between them was 25cm and so the distance between them after 1s is 25 - 10 = 15cm.

Yes, for projectiles, you normally consider the horizontal motion separately from the vertical motion as the horizontal component is constant velocity and the vertical motion is constant acceleration due to gravity.

0

reply

Thank you

(Original post by

X is projected towards Y. Its horizontal velocity is 0.1m/s and so the distance it travels is vt = 0.1*1=0.1m = 10cm. The original distance between them was 25cm and so the distance between them after 1s is 25 - 10 = 15cm.

Yes, for projectiles, you normally consider the horizontal motion separately from the vertical motion as the horizontal component is constant velocity and the vertical motion is constant acceleration due to gravity.

**Teenie2**)

X is projected towards Y. Its horizontal velocity is 0.1m/s and so the distance it travels is vt = 0.1*1=0.1m = 10cm. The original distance between them was 25cm and so the distance between them after 1s is 25 - 10 = 15cm.

Yes, for projectiles, you normally consider the horizontal motion separately from the vertical motion as the horizontal component is constant velocity and the vertical motion is constant acceleration due to gravity.

0

reply

Report

#7

**Teenie2**)

X is projected towards Y. Its horizontal velocity is 0.1m/s and so the distance it travels is vt = 0.1*1=0.1m = 10cm. The original distance between them was 25cm and so the distance between them after 1s is 25 - 10 = 15cm.

Yes, for projectiles, you normally consider the horizontal motion separately from the vertical motion as the horizontal component is constant velocity and the vertical motion is constant acceleration due to gravity.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top