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Initial and final velocity confusion watch

1. this may sound silly but for projectile motion I have trouble deciding whether the initial or final velocity should be 0...

Eg, A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 30o from the horizontal with a velocity of 40 m/s. How far will it travel in the air?

Horizontal

u = 40*cos30
v=

Vertical

u = 40*sin30
v=

In such a question, would the final velocity of the horizontal or vertical component be 0?
And also, is there a general guideline for me to follow so as to decide this?
2. (Original post by princejan7)
this may sound silly but for projectile motion I have trouble deciding whether the initial or final velocity should be 0...

Eg, A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 30o from the horizontal with a velocity of 40 m/s. How far will it travel in the air?

Horizontal

u = 40*cos30
v=

Vertical

u = 40*sin30
v=

In such a question, would the final velocity of the horizontal or vertical component be 0?
And also, is there a general guideline for me to follow so as to decide this?
The horizontal component remains constant throughout its flight. The vertical component reaches a velocity of 0 at the very top of its path, due to gravity, before coming back down.
3. As dknt said in effect, neither the initial nor the final velocity is zero. What is zero at the end of the home run, is the vertical displacement since it's come back to earth.
4. (Original post by dknt)
The horizontal component remains constant throughout its flight. The vertical component reaches a velocity of 0 at the very top of its path, due to gravity, before coming back down.
so the horizontal final velocity is also ucos40?

And what is it about the ""suvat"" equations of motion that only make them applicable for 1 dimension problems?
5. (Original post by princejan7)
so the horizontal final velocity is also ucos40?
yes. Initial and final.

And what is it about the ""suvat"" equations of motion that only make them applicable for 1 dimension problems?
The way the equations are formulated is such that the values of u,v,s,a (and t) can only be expressed as plus or minus, meaning that there is only one dimension present. In other words, they can only be up/down or left/right, for example. (Time forwards or backwards!)
They can, however, be applied to more than one dimension by considering the other dimensions separately.
Any two dimensional problem can be broken down into two (one dimensional) coordinate systems at right angles to each other.
The beauty in doing this is that in the case of projectile motion, the gravitational force acts vertically and has no effect on the horizontal motion. So you can analyse the vertical and horizontal motion separately. The horizontal component of velocity in this example remains constant, while the vertical component is affected by gravity.
Because of this, the vertical motion can be calculated using the suvat equations, and the horizontal using simple constant velocity.
The result is that complex motion can be analysed by breaking it up into components mutually at right angles, and treating each component separately.
6. (Original post by Stonebridge)
yes. Initial and final.

The way the equations are formulated is such that the values of u,v,s,a (and t) can only be expressed as plus or minus, meaning that there is only one dimension present. In other words, they can only be up/down or left/right, for example. (Time forwards or backwards!)
They can, however, be applied to more than one dimension by considering the other dimensions separately.
Any two dimensional problem can be broken down into two (one dimensional) coordinate systems at right angles to each other.
The beauty in doing this is that in the case of projectile motion, the gravitational force acts vertically and has no effect on the horizontal motion. So you can analyse the vertical and horizontal motion separately. The horizontal component of velocity in this example remains constant, while the vertical component is affected by gravity.
Because of this, the vertical motion can be calculated using the suvat equations, and the horizontal using simple constant velocity.
The result is that complex motion can be analysed by breaking it up into components mutually at right angles, and treating each component separately.
Thanks!!

One last thing, why is it that the "suvat" equations cant be used for the horizontal component?
And also, why is it that a ball thrown vertically is going to stay in the air the same amount of time as a ball thrown at an angle? I know its because gravity does not act on the horizontal component but that dosent make much sense to me at the moment..
7. (Original post by princejan7)
Thanks!!

One last thing, why is it that the "suvat" equations cant be used for the horizontal component?
And also, why is it that a ball thrown vertically is going to stay in the air the same amount of time as a ball thrown at an angle? I know its because gravity does not act on the horizontal component but that dosent make much sense to me at the moment..
because in both scenario, there is a uniform/constant force acting on both bodies..Hence , they accelerate at an equal rate---9.8m/s2 to be specific.
8. (Original post by princejan7)
Thanks!!

One last thing, why is it that the "suvat" equations cant be used for the horizontal component?
And also, why is it that a ball thrown vertically is going to stay in the air the same amount of time as a ball thrown at an angle? I know its because gravity does not act on the horizontal component but that doesnt make much sense to me at the moment..
The suvat equations are for uniformly accelerated motion. The "a" in those equations is this acceleration. If the object is not accelerating, you don't need suvat because a=0.
For example in v=u+at if the object is moving with uniform velocity, no acceleration, putting a=0 gives
v=u
Initial velocity equals final velocity. No acceleration. Uniform motion.
Then you just use velocity=displacement/time

In answer to your second question, a ball thrown vertically does not necessarily stay in the air the same time of as a ball thrown at an angle. It depends on the initial speed and the angle, for a start. It will only happen if the vertical component of the ball thrown at an angle, is the same as the actual speed the vertical ball was thrown upwards with.
For example, if you throw a ball vertically upwards at 10m/s it will stay in the air for 2 seconds. [use v=u+at and g=10m/s/s]
In order for a ball thrown at an angel to do the same, its vertical component must be 10m/s.
You could do this in any number of ways. If the angle to the horizontal is θ, and you throw it with velocity V, the vertical component is Vsinθ.
So it would work, for example, with a speed of 20m/s at an angle of 30 degrees.
Sin 30 = 0.5 and 20 sin 30=10.
But it would also work at other speeds and other angles so long as Vsinθ is 10. [eg a speed of 14.14m/s at an angle of 45 deg would also work. Check it out.]
9. (Original post by Stonebridge)
The suvat equations are for uniformly accelerated motion. The "a" in those equations is this acceleration. If the object is not accelerating, you don't need suvat because a=0.
For example in v=u+at if the object is moving with uniform velocity, no acceleration, putting a=0 gives
v=u
Initial velocity equals final velocity. No acceleration. Uniform motion.
Then you just use velocity=displacement/time

In answer to your second question, a ball thrown vertically does not necessarily stay in the air the same time of as a ball thrown at an angle. It depends on the initial speed and the angle, for a start. It will only happen if the vertical component of the ball thrown at an angle, is the same as the actual speed the vertical ball was thrown upwards with.
For example, if you throw a ball vertically upwards at 10m/s it will stay in the air for 2 seconds. [use v=u+at and g=10m/s/s]
In order for a ball thrown at an angel to do the same, its vertical component must be 10m/s.
You could do this in any number of ways. If the angle to the horizontal is θ, and you throw it with velocity V, the vertical component is Vsinθ.
So it would work, for example, with a speed of 20m/s at an angle of 30 degrees.
Sin 30 = 0.5 and 20 sin 30=10.
But it would also work at other speeds and other angles so long as Vsinθ is 10. [eg a speed of 14.14m/s at an angle of 45 deg would also work. Check it out.]
thanks a lot!

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