Projectile Motion: Calculating Landing Position and Velocity Components

AI Thread Summary
A ball rolls off a 1.5m high table at a constant speed of 5m/s, leading to questions about its landing position and velocity components. The initial vertical velocity (Vo)y is zero since the ball rolls horizontally. Using kinematic equations, the time the ball is in the air is calculated to be approximately 0.5533 seconds, resulting in a horizontal landing distance of about 2.8 meters. The final horizontal velocity (Vx) remains 5m/s, while the final vertical velocity (Vy) can be determined using the equation V_fy = V_oy + at. The discussion emphasizes the application of independent motion equations for horizontal and vertical components in projectile motion.
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Hello hello, i have a question for you tonight:

a ball on a 1.5m high table moving at a constant 5m/s rolls of the table.

1. where does it land
2. what are the Vx and Vy values JUST before it hits the ground
thanks so much!

Homework Statement


Homework Equations


The Attempt at a Solution



@pbdude: I am sorry..i didnt really get what you said and i made a topic...thanks for helping though :)
 
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Hey there! I will ask you similar questions as the other thread to start you off:

Since we know that it rolls off horizontally, what can you say about the initial velocity in the Y-direction? That is, what is (Vo)y ?

Also, what are the relevant kinematic equations?
 
Saladsamurai said:
Hey there! I will ask you similar questions as the other thread to start you off:

Since we know that it rolls off horizontally, what can you say about the initial velocity in the Y-direction? That is, what is (Vo)y ?

Also, what are the relevant kinematic equations?


okay, (Vo)y=0 because its only going forward.

other formulas?

Vx=Vcos(theta)
Vy=Vsin(theta)
Dx=Vx(t)
Range/(Dy?)=V²(sin2(theta))/9.8
Dy=Vy(t) + .5at²
 
Okay, Don't worry about the 'range' formula. Let's stick to just:

Dy=Vy(t) + .5at²
Vf=Vo + at
Vf2=Vo2 + 2a(df-do)

Okay?

We also know ay=g=-9.81 m/s2

So, can we apply EQ 1 in the y-direction to find out how long the ball was in the air for?
 
Last edited:
okay, i can do that but i don't know EQ 3 :(
Dy=Vy(t) + .5at²
1.5=0+4.9t² (my teacher uses 9.8 m/s²)

sooo t=.5533?
 
so, i hope double posting is allowed but
the EQ Vf= Vo + at means that Vf=Vxf or Vyf, and Vo=Vox or Voy?
 
Now how far did it land from the table (the x-direction)?

You have Vox, and you have the amount of time it travels at that speed. You should be able to answer part (1) now.

the EQ Vf= Vo + at means that Vf=Vxf or Vyf, and Vo=Vox or Voy

Yes. I gave you the 'general' form of the EQs. You can apply them to each direction independently as needed.
 
Saladsamurai said:
Now how far did it land from the table (the x-direction)?

You have Vox, and you have the amount of time it travels at that speed. You should be able to answer part (1) now.

OHHH

right :P

5m/s for Vox and .5533 for t

sooo 2.766 M and the answer in the book is 2.8 OH YEAH


thanks SO much :)
 
so how do i find Vx and Vy? if i don't know theta?
 
  • #10
Great! For part (2) you know that Vox is constant, so V_fx is the same.

To find V_fy, apply the kinematic EQ: V_fy= V_oy + at

where a=g and 't' you have found.
 
  • #11
Saladsamurai said:
Great! For part (2) you know that Vox is constant, so V_fx is the same.

To find V_fy, apply the kinematic EQ: V_fy= V_oy + at

where a=g and 't' you have found.

ahhh thanks again :D
 
  • #12
Anytime :smile:
 
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