Can Initial Speeds of 0m/s Solve This Kinematics Problem?

[Ukitake Jyuushirou]

Alice, Bob and Daniel are standing on the edge of a canyon which is 100m deep and 20m wide. On the opposite side of the canyon there is a 10m high cave, 15m below the top of the cliff. They each throw a rock horizontally with the following initial speed to see who can get it into the cave.

Alice rock speed 5m/s
Bob rock speed 10m/s
Daniel rock speed 20m/s

for this qn, i do not understand why initial velocity is 0...hmmm ? becoz only with the initial speed as 0 can i work out the correct ans...if i were to use the values given, i will get a wrong ans...

 

[arunbg]

The initial velocity in the vertical direction is 0 since the stones are being thrown horizontally.
In the horizontal direction, the stone always has the velocity as given in the question .

 

[Hootenanny]

In the horizontal direction, the stone always has the velocity as given in the question .

Ignoring air resistance of course

~H

 

[arunbg]

And various other factors, of course

 

[Andrew Mason]

Alice, Bob and Daniel are standing on the edge of a canyon which is 100m deep and 20m wide. On the opposite side of the canyon there is a 10m high cave, 15m below the top of the cliff. They each throw a rock horizontally with the following initial speed to see who can get it into the cave.

Alice rock speed 5m/s
Bob rock speed 10m/s
Daniel rock speed 20m/s

for this qn, i do not understand why initial velocity is 0...hmmm ? becoz only with the initial speed as 0 can i work out the correct ans...if i were to use the values given, i will get a wrong ans...

Work out the equation for speed in terms of the vertical drop distance over the 20 m. width. Then work out the range of permitted speeds that would result in a drop of between 5 and 15 m. (I assume the floor of the cave is 15 m. below the top).

AM

 

[Ukitake Jyuushirou]

The initial velocity in the vertical direction is 0 since the stones are being thrown horizontally.
In the horizontal direction, the stone always has the velocity as given in the question .

hmmm...but the diagram that is drawn on the paper shows that when they throw the rock, it will form an angle...since they are at the top of the cliff, the cave is opposite them and slightly below them...

when solving this sort of qn, do i have to factor in the angle? or i can juz treat the qn independently in terms of x and y axis

 

[Hootenanny]

hmmm...but the diagram that is drawn on the paper shows that when they throw the rock, it will form an angle...since they are at the top of the cliff, the cave is opposite them and slightly below them...

when solving this sort of qn, do i have to factor in the angle? or i can juz treat the qn independently in terms of x and y axis

You would need factor in the angles when resolving the horiztonal and vertical components. For example, if a ball is thrown with an intial velocity of 10 m.s^-1 at an angle of 60 degrees above the honizontal, then the vertical and horizontal velocities would be as follows;

$$V_{x} = 10\cos 60$$

$$V_{y} = 10\sin 60$$

Do you follow?

~H

 

[Ukitake Jyuushirou]

You would need factor in the angles when resolving the horiztonal and vertical components. For example, if a ball is thrown with an intial velocity of 10 m.s^-1 at an angle of 60 degrees above the honizontal, then the vertical and horizontal velocities would be as follows;

$$V_{x} = 10\cos 60$$

$$V_{y} = 10\sin 60$$

Do you follow?

~H

sort of...so u mean if let's say when the qn gives us the angle and the velocity then i would have to factor it in...otherwise i can ignore the angle in this qn.

but if there is no vertical velocity...and the horizontal velocity doesn't play a part in getting the rock into the cave, why does the qn still give the horizontal velocity?!

 

[Hootenanny]

RE: AM's post. You need to work out a range of initial velocities. As the intial velocity is horizontal, there is no angle. Consider the horizontal (x) and vertical (y) components separately.

Horizontally

You know that horizontally the rock must travel 20m (the width of the canyon). Ignoring air resistance etc., this velocity is constant, therefore the intial velocity is given by;

$$V_{x} = \frac{20}{t}$$

Vertically

You that vertically the shortest distance the rock must fall is 5m (I'm assuming the 15m is measured to the base of the cave as I cannot see your diagram). Ignore air resistance, the only acceleration acting in the y plane is gravity, therefore using kinematic equations the minimum time taken for the rock to fall 5m is given by;

$$t = \sqrt{\frac{2s}{a}}$$

Now, if you input this time into the equation for horizontal velocity, this will give you the maximum intial velocity.

Next you will need to work out the maximum time taken for the rock to fall vertically (15m), then input this time into the Vx equation to obtain the minimum intial velocity.

Do you follow?

~H

 

[Ukitake Jyuushirou]

diagram ...

 

[Hootenanny]

Thank you for the diagram Ukitake, however, it will probably take a while for the mentors to approve this attachment. Do you understand what I was saying in my previous post?

~H

 

[Ukitake Jyuushirou]

sorry...my internet was down for a long time

in light of the diagram i have drawn, does wat u stated still hold true?

 

[Hootenanny]

sorry...my internet was down for a long time

in light of the diagram i have drawn, does wat u stated still hold true?

No problem my friend. I cannot say, because as you can see the attachment is still pending. If it isn't approved soon I will PM one of the mentors to see what can be done.

~H

 

[Ukitake Jyuushirou]

No problem my friend. I cannot say, because as you can see the attachment is still pending. If it isn't approved soon I will PM one of the mentors to see what can be done.

~H

thanks :) :)

 

[Hootenanny]

Okay, in light of your diagram the vertical range requires alteration, however the horizontal component still holds (the rock still must travel a horizontal distance of 20m).

Vertically

Vertically, the minimum distance the rock must fall to enter the cave is 15m and the maximum distance is 25m. So we can now apply equations of uniform acceleration to find the flight time. The equation required is s = ut + ¹/₂at². As the rocks are thrown horizontally the intial vertical velocity is zero, thus the "ut" term drops out leaving s = ¹/₂at². Rearranging to find t gives;

$$t = \sqrt{\frac{2s}{a}}$$

Next you need to substitute in the maximum and minimum values for displacement (s) to obtain the maximum and minimum values for flight time. After obtaining the times you need to substitute them into the equation for horizontal motion to obtain the minimum and maximum intial horizontal velocities. Any velocity which lies between these two values will result in the rock being thrown into the cave.

Do you follow?

~H

 

[arunbg]

I don't see any reference to angles in the diagram.
It's a straightforward question .
If you still have difficulties, feel free to post them .

 

[Ukitake Jyuushirou]

Okay, in light of your diagram the vertical range requires alteration, however the horizontal component still holds (the rock still must travel a horizontal distance of 20m).

Vertically

Vertically, the minimum distance the rock must fall to enter the cave is 15m and the maximum distance is 25m. So we can now apply equations of uniform acceleration to find the flight time. The equation required is s = ut + ¹/₂at². As the rocks are thrown horizontally the intial vertical velocity is zero, thus the "ut" term drops out leaving s = ¹/₂at². Rearranging to find t gives;

$$t = \sqrt{\frac{2s}{a}}$$

Next you need to substitute in the maximum and minimum values for displacement (s) to obtain the maximum and minimum values for flight time. After obtaining the times you need to substitute them into the equation for horizontal motion to obtain the minimum and maximum intial horizontal velocities. Any velocity which lies between these two values will result in the rock being thrown into the cave.

Do you follow?

~H

in order to work out the velocity i need acceleration and since acceleration of horizontal plane is 0. that keeps screwing my answers

 

[Ukitake Jyuushirou]

I don't see any reference to angles in the diagram.
It's a straightforward question .
If you still have difficulties, feel free to post them .

yea but i have problem with the concepts of kinematics...

we're taught to treat x component separately from y component. i am given initial velocity for horizontal component, we know gravity (y component) is 9.8, i can work out the time taken for the rock to fly horizontally but since the cave is located in the y-axis i cannot use the time, initial velocity...

 

[Hootenanny]

in order to work out the velocity i need acceleration and since acceleration of horizontal plane is 0. that keeps screwing my answers

The acceleration in the horizontal plane is zero.

However, the acceleration in the vertical plane is _____ m.s^-2?

Consider the horizontal and vertical components completely sperately.

~H

 

[Ukitake Jyuushirou]

The acceleration in the horizontal plane is zero.

However, the acceleration in the vertical plane is _____ m.s^-2?

~H

9.8

but i can't use a vector for vertical plane on the horizontal plane...or can i?

 

[Hootenanny]

9.8

but i can't use a vector for vertical plane on the horizontal plane...or can i?

No. But you can work out the time taken for the particle do fall the required distance. Have you read my posts?

~H

 

[Ukitake Jyuushirou]

No. But you can work out the time taken for the particle do fall the required distance. Have you read my posts?

~H

yes

"Next you need to substitute in the maximum and minimum values for displacement (s) to obtain the maximum and minimum values for flight time. After obtaining the times you need to substitute them into the equation for horizontal motion to obtain the minimum and maximum intial horizontal velocities. Any velocity which lies between these two values will result in the rock being thrown into the cave."

i can work out the max and minimum times for flight which is 2.25 and 1.74 respectively. i am a little confuse about the 2nd part of the paragraph esp on the "minimum and maximum inital horizontal velocities"...

 

[Hootenanny]

i can work out the max and minimum times for flight which is 2.25 and 1.74 respectively. i am a little confuse about the 2nd part of the paragraph esp on the "minimum and maximum inital horizontal velocities"...

Okay. Now if we consider the horizontal plane only; there are no forces acting (we are neglecting air resistance etc) so there is no acceleration thus the rock will be traveling at a constant horizontal velocity. Know as you know the equation for constant velocity is;

$$v = \frac{\Delta s}{\Delta t}$$

Now, we need to find the velocities that will allow the stone to travel the 20m into the cave in the time taken for the stone to fall the required distance - the times you found above. Do you follow?

~H

 

[Ukitake Jyuushirou]

Okay. Now if we consider the horizontal plane only; there are no forces acting (we are neglecting air resistance etc) so there is no acceleration thus the rock will be traveling at a constant horizontal velocity. Know as you know the equation for constant velocity is;

$$v = \frac{\Delta s}{\Delta t}$$

Now, we need to find the velocities that will allow the stone to travel the 20m into the cave in the time taken for the stone to fall the required distance - the times you found above. Do you follow?

~H

yes! with this i can find the correct ans...thanks

but the time taken is for the stone to fall in the vertical axis, i can use it to find velocity? in the horizontal axis? becoz in my textbook, they separate this type of qn into x and y component...

certain vectors are only exclusive to either x or y components...is time an exclusive component?

 

[Hootenanny]

Yes, time is constant in both components. Think of it this way, a runner sets of at a constant velocity of 5m.s^-1 east. A car starts from the same point at the same time from rest but accelerates at a rate of 10m.s^-2 south. After 5 seconds what is the velocity of each? What is the displacement?

This exactly the same principle.

~H

 

[Ukitake Jyuushirou]

Yes, time is constant in both components. Think of it this way, a runner sets of at a constant velocity of 5m.s^-1 east. A car starts from the same point at the same time from rest but accelerates at a rate of 10m.s^-2 south. After 5 seconds what is the velocity of each? What is the displacement?

This exactly the same principle.

~H

ok...becoz all along i tot, there is a diff time for each component and each are exclusive so i can't use the time for vertical on the horizontal component...but now i see...thanks everyone for all the help

 

[Hootenanny]

ok...becoz all along i tot, there is a diff time for each component and each are exclusive so i can't use the time for vertical on the horizontal component...but now i see...thanks everyone for all the help

I think we've just had a break through . No problem, it was a pleasure.

~H

 

[Ukitake Jyuushirou]

a girl drops a water filled balloon to the ground, 6m below. If the balloon is released from rest, how long is it in the air?

V (ini) = 0
V (final) = ?
a = -9.8
x = 6

V (final) = V (ini) + 2ax
V (final) = 10.8

at + V (ini) = V
t = -1.1s

- 1.1 is the answer according to my workings but u can have negative time...rite?

 

[Hootenanny]

V (final) = V (ini) + 2ax

The expression is actually $V_{f}^{{\color{red}2}} = V_{i}^{{\color{red}2}} + 2ax$. However, your answer of 1.1 is correct. It is possible to get a negative time (this is due to the symmetry of the quadratic curve) however, the positive time is usually taken as the answer. An easier expression to use would be $x = v_{i}t + \frac{1}{2}at^2$.

 

[Ukitake Jyuushirou]

The expression is actually $V_{f}^{{\color{red}2}} = V_{i}^{{\color{red}2}} + 2ax$. However, your answer of 1.1 is correct. It is possible to get a negative time (this is due to the symmetry of the quadratic curve) however, the positive time is usually taken as the answer. An easier expression to use would be $x = v_{i}t + \frac{1}{2}at^2$.

sweet! thanks :)