If two balls, with identical masses are launched in the air at different angles (one at 15 degrees the other at 30 degrees): would they have the same impact velocity? Will they have different maximum heights and different horizontal distances? Thanks!
Kinetic Energy and Projectile Motion
- Thread starter jgalt42
- Start date
- Joined
- Dec 1, 2011
- Messages
- 18,579
- Reaction score
- 57
v2=vo2+2ax So different impact velocities.
y=voyt+1/2at^2 So different max heights.
x=voxt So different horizontal ranges.
- Joined
- Nov 1, 2011
- Messages
- 175
- Reaction score
- 2
If you're trying to think about stuff like this conceptually it helps to use extreme examples. Say you launch ball A at an angle of 90 degrees and ball B at an angle of 5 degrees. You launch them with the same initial velocity. The Y component of ball A is going to be much greater than the Y component of ball B. All the force is going to driving the ball up with ball A where as it is mostly driving the ball forward with ball B. This will cause ball A to go higher than ball B. Since ball A is going higher than ball B it will take longer to fall, have more time to accelerate and thus have greater impact velocity. Remember that velocity = 0 at the top of the flight. You can use the kinematics equations to solve for specifics. Just remember to separate into components.
Edit: My bad, meant to say greater Y component not impact velocity. A has greater Y component and B has greater X component. Combine X and Y components to find impact velocity.
Edit: My bad, meant to say greater Y component not impact velocity. A has greater Y component and B has greater X component. Combine X and Y components to find impact velocity.
Last edited:
- Joined
- Dec 30, 2009
- Messages
- 2,641
- Reaction score
- 527
All said is true. What will be the same is the impact speeds (or the magnitude of the impact velocities). Not that the question asked for that but something to keep in mind.
---
I am here: http://maps.google.com/maps?ll=47.652977,-122.308263
---
I am here: http://maps.google.com/maps?ll=47.652977,-122.308263
- Joined
- Dec 1, 2011
- Messages
- 18,579
- Reaction score
- 57
Also remember that for identical objects with identical initial velocity, 45 degrees will give you the greatest range, 90 degrees the greatest height.
Complementary angles will have the same range, but different heights.
Complementary angles will have the same range, but different heights.
- Joined
- Dec 1, 2011
- Messages
- 18,579
- Reaction score
- 57
If you take into consideration the x and y components then the angles matter. If you're just doing simple conservation of energy, then it won't always matter.
For example, an object is launched at a 30deg angle to the horizontal with velocity 5m/s.
Since energy is conserved, the energy of the object initially must be the same as when it is at its peak.
1/2mv^2=mgh
v^2=2gh
v=sqrt(2gh)
- Joined
- Nov 1, 2011
- Messages
- 175
- Reaction score
- 2
He's talking about greatest horizontal distance. Say you throw a ball at 30, 45, 60, 70, 80 degrees ect, all at the same initial velocity, the one at 45 degrees will go the furthest.
- Joined
- Dec 1, 2011
- Messages
- 18,579
- Reaction score
- 57
Yea, range = horizontal distance. Notice that the rule about the 45deg is only true when deltay=0. In other words, when the object takes off and lands at the same height.
Complementary angles will have opposite vertical and horizontal velocity components. If one is 10 vertical and 5 horizontal, the other is 10 horizontal and 5 vertical. So make a triangle with these sides. The hypotenuse will be the same!
- Joined
- Dec 30, 2009
- Messages
- 2,641
- Reaction score
- 527
By complimentary angles he means something like α and 90-α, for example 10 and 80 degrees. At both of these angles the object will reach the same distance horizontally but in the first case its maximum height will be much lower than the second.
The conceptual explanation is that at low angles you're throwing the object so close to the ground that it falls too quick and does not have time to go far. At high angles you throw it mostly upwards so you don't get enough motion along the horizontal.
The math version for those inclined to read it: Let's say that you're throwing at angle α with the horizon and speed v. The horizontal distance is d=v^2 * sin(2α
/g. The maximum vertical distance is (v.sin(α)^2/g. You'll get maximum distance for sin(2α
=1 or 2α=90 or α=45. For other angles, we have sin(2*(90-α)=sin(180-2α=sin(2α which gives you same distance for α and 90-α.For all angles between 0 and 90 the height is continuously increasing - the higher the angle, the higher the max height is.
- Joined
- Dec 30, 2009
- Messages
- 2,641
- Reaction score
- 527
But the hypotenuse is not the maximum distance. What you say is correct but it is also correct for the case when the object lands at higher/lower height when the distances are not the same for complementary angles.
- Joined
- Mar 14, 2007
- Messages
- 118
- Reaction score
- 7
Say I get a question where I am given a mass of an object as well as initial velocity and maximum hight.
Could I calculate the initial kinetic energy then subtract the potential energy at the top to get the horizontal velocity at the top?
Shouldn't (if the ball is launched and impacted on the same level surface) the ball have the same velocity magnitude upon impact as it initially did regardless of the angle it was launched from due to conservation of energy?
If you launch a ball at 5 degrees and one at 30 degrees either way the upward velocity is converted to potential energy and then back to kinetic energy. If you launch them both at 30ms/s shouldn't they both land on a level surface at 30m/s?
Could I calculate the initial kinetic energy then subtract the potential energy at the top to get the horizontal velocity at the top?
Shouldn't (if the ball is launched and impacted on the same level surface) the ball have the same velocity magnitude upon impact as it initially did regardless of the angle it was launched from due to conservation of energy?
If you launch a ball at 5 degrees and one at 30 degrees either way the upward velocity is converted to potential energy and then back to kinetic energy. If you launch them both at 30ms/s shouldn't they both land on a level surface at 30m/s?
- Joined
- Dec 1, 2011
- Messages
- 18,579
- Reaction score
- 57
can you please explain how complementary angles will have same range but different heights?[/QUOTE]
The range equation is vo^2sin2theta/g. Lets use 30deg and 60deg as our complementary angles.
What is sin60deg? 0.867
What is sin120deg? 0.867
Milski has a good explanation of why, but if you need proof, all you need to do is plug in complementary angles and the equation gives you the same answer for both.
- Joined
- Dec 1, 2011
- Messages
- 18,579
- Reaction score
- 57
Assuming constant acceleration, you don't need any information other than the initial velocity to find the horizontal velocity at any given point or time. The initial horizontal velocity is equal to the horizontal velocity at all points, barring acceleration. The equations we are referring to are only valid with constant acceleration anyway, so for the purposes of MCAT, horizontal velocity in projectile motion is constant. vox=vx.
