projectiles

[annjolino]

1. The problem statement, all variables and given/known data
throws a small rock at a beehive from the ground at an angle θ above the horizontal. throws it at velocity v. the rock strikes the beehive just as it reaches the top of its trajectory.
a)ignoring air resistance, what is the speed with which the rock hits thee beehive
b)calculate the speed v=18.0ms-1 θ=55*
c) how high is the beehive



2. Relevant equations
Vy=1/2gt
Vx???


3. The attempt at a solution

ok i got the time it took to reach its max height was
18.0ms-1/9.8ms-1=1.8s
not sure if that was right equation but it looks like the right answer

a) the Vy at the top of trajectory would equal 0 due to the downward acceleration of gravity
and the Vx i am not sure on.

b)i tried to use trig in a way to solve for just the equation and i got this
i started with finding out the distance travelled along the rocks path
d=1/2gt^2
d=1/2(9.8ms-1)(1.8s^2)
d=15.88m

i then factored in the angle
55*=cos(x/15.88m)
x=cos55*(15.88)
x=9.11ms-1
i dont know if this is right..

the height i think may actually be what i calculated for a) 15.88m

[tiny-tim]

1. The problem statement, all variables and given/known data
throws a small rock at a beehive from the ground at an angle θ above the horizontal. throws it at velocity v. the rock strikes the beehive just as it reaches the top of its trajectory.
a)ignoring air resistance, what is the speed with which the rock hits thee beehive
b)calculate the speed v=18.0ms-1 θ=55*
c) how high is the beehive
…
a) the Vy at the top of trajectory would equal 0 due to the downward acceleration of gravity
and the Vx i am not sure on.

Hi annjolino! :smile:

(are you the same person as lilmissbossy?)

You get the answer to a) from the fact that there are no horizontal forces … so the change in the horizontal component of velocity is … ? :smile:

Have you done "KE + PE = constant" in class? If so …

Hint: easy way: you know the horizontal component of velocity from a).

So the KE at the highest point is … ?

Then use KE + PE = constant! :smile:

[annjolino]

well actually my course is online uni, so i basically have to teach myself and i have found it quite difficult... so even if the rock is thrown at an angle it does not have a horizontal force?
i am confused if the rock hits the beehive at the top of its trajectory which i believe will have a vertical speed of 0ms. The question ask what speed will it hit at?? so i assumed that it would have at least a horizontal speed??? i understand that the horizontal speed does not change like the vertical velocity... this is all too confusing

[mite]

well actually my course is online uni, so i basically have to teach myself and i have found it quite difficult... so even if the rock is thrown at an angle it does not have a horizontal force?
i am confused if the rock hits the beehive at the top of its trajectory which i believe will have a vertical speed of 0ms. The question ask what speed will it hit at?? so i assumed that it would have at least a horizontal speed??? i understand that the horizontal speed does not change like the vertical velocity... this is all too confusing

The rock is thrown at an angle isn't it then the initial velocity can be resolved into two components one along the horizontal direction and another along vertical direction right

As there is force along the vertical direction (in opposite direction to motion hence decelerates) vertical component of intial velocity changes.

It has only kinetic energy in the begining. As its height increases its kinetic energy decreases & its potential energy increases but the total energy is the same isn't it (as we are not taking into account the air resistance) When it reaches the heighest point it has only potential energy no kinetic energy So this potential energy at this heighest point is equal to the total energy Using this the speed with which it hits the beehive can be calculated. This speed is just the horizontal component of the initial velocity.

As there is no force along horizontal direction there is no change in velocity along this direction that is horizontal component of initial velocity does not change

[tiny-tim]

... so even if the rock is thrown at an angle it does not have a horizontal force?

The "throwing" force is partly horizontal, yes, and that is what gives it its initial horizontal velocity.

But the "throwing" force is over as soon as you let go … from then on, there's no horizontal force, so no reason for the initial horizontal velocity to change. :smile:
i understand that the horizontal speed does not change like the vertical velocity... this is all too confusing

Yes, you're right … horizontal velocity fixed, vertical velocity changing.

That's because horizontal force zero, vertical force non-zero … as mite says:
As there is no force along horizontal direction there is no change in velocity along this direction that is horizontal component of initial velocity does not change
:smile:

[annjolino]

Ok i understand that, so the initial velocity when thrown at an angle is used to find both the Vx and Vy components... so then does that mean that the Vy+Vx=Vi ??

[tiny-tim]

Ok i understand that, so the initial velocity when thrown at an angle is used to find both the Vx and Vy components... so then does that mean that the Vy+Vx=Vi ??

hi annjolino! :smile:

If you mean that as a vector equation, then yes. :smile:

The initial velocity vector Vi is the sum of Vx in the x direction plus Vy in the y direction

[annjolino]

ok so i am not sure how to find one of them with the given information... as you can see from the initial msg i tried one way using something that i think i made up in my head...

[Chrisas]

You are given an initial velocity (as in a vector) with the magnitude and direction (angle) given. From that vector, you can compute both the starting, or initial, components, Vx and Vy.

You are asked for the "speed", which is the magnitude of the velocity, at the top of the trajectory. The phrase 'at the top of the trajectory' should tell you what the final Vy is. All you need to compute the speed is the final Vx. The fact that no forces act in the horizontal direction should tell you how the initial Vx changes into the final Vx.

[tiny-tim]

It's not clear from your original post …

did you mean that the question gives the initial velocity Vi as 18m/s at an angle of 55º?

If so, the the components of Vi are 18 cos55º and 18 sin55º …

draw a triangle to see which is Vx and which is Vy. :smile:

[annjolino]

ok i see where i was going wrong,
i have
Vx= 18cos 55* = 10.32
Vy= 18sin 55* = 14.74

i get confused when i try to use trig and velocity at the same time i never know what to use...

so the speed at which the rock hits is 10.32ms in the horizontal component a 0ms in the vertical component.

is that right???

[tiny-tim]

so the speed at which the rock hits is 10.32ms in the horizontal component a 0ms in the vertical component.

is that right???

Yes!!! :smile:

Can you work out the height now (using KE + PE = constant)?

[annjolino]

i know that PE=mgh and KE=1/2mv^2.... i have just looked for this is my text and i can see an example that shows the relationship between them in respect to a free fall, but i cannot seem to find much else....

[annjolino]

could i use something like finding the total flight time and multiply that by the vertical component of the initial velocity???

[annjolino]

i hope i have the right idea here

i divided the Vy by 9.8ms to find the time it would reach its max height...
14.74/9.8 = 1.5s

i then multiplied the Vy by the 1.5s...
14.74msx1.5s= 22.11m
??????????

[tiny-tim]

i divided the Vy by 9.8ms to find the time it would reach its max height...
14.74/9.8 = 1.5s

That's right! :smile:

Unfortunately, the question doesn't ask you for the time. :rolleyes:
i then multiplied the Vy by the 1.5s...

Nooo … that only works for constant speed.

(Though there is a way you could adjust it … :wink:)

Let's get back to the simplest method:
i know that PE=mgh and KE=1/2mv^2.... i have just looked for this is my text and i can see an example that shows the relationship between them in respect to a free fall, but i cannot seem to find much else....

Forget examples … just work it out from the equation KE + PE = constant.

(in the exam, you won't have the examples in your head, but you will have that basic equation in your head! :wink:)

What is KE initial?

What is KE final?

What is PE final minus PE initial?

And then … ? :smile:

[annjolino]

the truth i have absolutely no idea, i have never used that equation. if i am supposed to factor in the values for KE and PE how do i find them if i only have the velocity and no mass?????

[tiny-tim]

the truth i have absolutely no idea, i have never used that equation. if i am supposed to factor in the values for KE and PE how do i find them if i only have the velocity and no mass?????

'cos they both have the mass, m, so you just divide everything by m …

KE/m + PE/m = constant. :wink:

[annjolino]

Ok bear with me here i could very well have stuffed this up......

PE+KE=Constant
PE/m + KE/m = constant
mgh/m + 1/2mv^2/m = constant
cancel out m
gh+v^2 = constant
gh=-v^2
h=-v^2/g

does that look right if it does i got 33.06m

that seems a little high???

[annjolino]

i thnk i left out the 1/2 in the equation after i cancelled out the m should i have left it there to say

gh+1/2v^2 ????
that gives me a more believable answer of 16.5m

[Banaticus]

We know that \frac{\triangle V}{a}=t and that the initial velocity in the vertical direction was about 14.74 m/s and that the final velocity was 0, that a was 9.8m/s, we can say that t is about 1.5s.

We also know that \triangle X=\frac{1}{2}(V+V_0)t, so the change in X is given by 1/2 of the change in velocity multiplied by the time, when under a constant acceleration (and the force of gravity is fairly constant). So, the height was just over 11 feet.

But, how tall was the guy who threw it? Surely we should add an extra six feet or so for how high off the ground his hand was when he threw it? Maybe four feet if he threw it underhanded.

[tiny-tim]

i thnk i left out the 1/2 in the equation after i cancelled out the m should i have left it there to say

gh+1/2v^2 ????
that gives me a more believable answer of 16.5m

No … you've only used the intial KE … you have to use the change in KE.

[annjolino]

i am really lost final KE = 0 and initial = 14.74 so change would equal-14.74?????

i am way past confused now..... :cry:

[Banaticus]

You don't "need" to calculate KE or PE for this problem. If you were trying to figure out the work done by friction or some other force in a given time or you had a changing acceleration because you're looking at a pendulum, then yeah you'd have to figure that out. But all we want is the the speed of the projectile and the height of the object and for that, all we need are the basic speed, velocity, acceleration and time formulas that I just used.
So, the height was just over 11 feet.

But, how tall was the guy who threw it? Surely we should add an extra six feet or so for how high off the ground his hand was when he threw it? Maybe four feet if he threw it underhanded.
Dang it, I meant that the height was just over 11 "meters", not feet. Whoops, I hate when that happens. Maybe add an extra meter or two to the height for the guy's height (because the ball certainly wasn't launched from ground level -- the problem said that the ball was thrown and you'd have to be down in a dugout or otherwise below ground to throw a ball from ground level). But that's a subject that you should take up with your physics teacher.

Taking significant figures into account, 11.1 meters should be the answer that you want.

[annjolino]

Thank you that feet really threw me off... i can understand what formula you were using i couldnt understand how to rearrange that KE+PE equation to get what i wanted..
no the question only states that it was thrown from the ground at the 55* angle so i dont thnk we have to factor any extra height in

Thankyou :smile:

[tiny-tim]

Hi annjolino! :smile:

(I've just got up … :zzz:)
i am really lost final KE = 0 …

No … final KE is based on the whole final velocity … so it's mVx2/2.

hmm … I think you were following what mite said … I think he was getting confused with what happens when you throw something straight up …
It has only kinetic energy in the begining. As its height increases its kinetic energy decreases & its potential energy increases but the total energy is the same isn't it (as we are not taking into account the air resistance)
When it reaches the heighest point it has only potential energy no kinetic energy So this potential energy at this heighest point is equal to the total energy
… but the rest of what he said is good. :smile:

So your KE + PE equation (divided by m :wink:) is (V2 - Vx2)/2 = gh …

which you could even shorten to Vy2/2 = gh. :smile:
i then multiplied the Vy by the 1.5s...
Nooo … that only works for constant speed.

(Though there is a way you could adjust it … :wink:)

Your method via t also works, and the way to adjust it is to notice that
distance = time x average speed
(for a constant acceleration, such as g) …

so instead of multiplying Vy by 1.5s, you could have multiplied Vy/2 by 1.5s (as Banaticus did) …

h = Vyt/2, t = Vy/g, so h = Vy2/2g …

which is the same as the KE + PE result.

The KE + PE method is quicker, since you don't spend time calculating t in the middle (and the longer you take, the more chance of making mistakes! :wink:)

[annjolino]

ok so this is what i got hopefully its the last for this.....

i am assuming that Vx is the velocity in the horizontal direction as we have said throughout the problem... so here goes

Vo^2 - Vx^2/2 = gh
18ms-10.32ms/2 = 9.8ms x h
108.7488/9.8ms = h
h= 11.1m

please let that be right....

[tiny-tim]

:biggrin: Woohoo! :biggrin:

[annjolino]

thank you so much i can make sense of this when its explained to me.... its a little tough to work out by myself.....:biggrin: