Some Phyiscs Questions

S

srothman

Expert Member
Joined
Mar 30, 2010
Messages
2,756
Reaction score
10
Location
Pretoria
Looking for some clarity on some basic physics scenarios, but because I'm a muppet, I need some explanations. I ducked out of high-school physics the moment I could, and I kind of regret it now...

Vertical Projectile Motion

A stone is thrown vertically upward at 20m.s-1 from a cliff 18m above the ground.
cliff.JPG

a) Determine the time it takes for the stone to pass the starting point in the way down.

Solution
v(i) = +20m.s^1
a = -9.8m.s^2
v(f) = -20m.s^1
Δt = ?

v(f) = v(i) + aΔt
-20 = +20 + (-9.8)Δt <<<<---- this is where the wheels come off completely....
Δt = 4.08s

b) Determine the maximum height above the cliff reached by the stone.

Solution
v(i) = +20m.s^1
a = -9.8m.s^2
v(f) = 0m.s^1
Δy = ?

v(f)^2 = v(i)^2 +2aΔy
(0)^2 = (20)^2 + 2(-9.8)Δy
Δy = +20.41m

c) Determine the maximum velocity of the stone when it is 11m above the foot of the cliff.

Solution
v(i) = +20m.s^1
a = -9.8m.s^2
v(f) = ?

v(f) = v(i) + 2aΔy
v(f) = (20)^2 + 2(-9.8)(-7)
v(f) = 23.18 m.s^1

d) Where, relative the cliff, is the stone when it reaches a velocity of 15m.s?

Solution
v(i) = +20m.s^1
a = -9.8m.s^2
v(f) = + or - 15m.s^1
Δy = ?

v(f)^2 = v(i)^2 + 2aΔy
(+- 15)^2 = (20)^2 + 2(-9.8)Δy
Δy = +8.93m

This is straight out of a handbook, but I can't, for the love of me, understand how they get to the solutions.

[video=youtube_share;AR6eXWNJzoY]https://youtu.be/AR6eXWNJzoY[/video]
 
Do you have the formulas for displacement?

I'll do one solution in a separate post, but I don't like spoonfeeding people and you've not really indicated what you don't understand.

I find with movement mechanics like this stuff, it helps to write down everything you have (a, s, v, etc) and to have a list of the formulas that you may need to use. It's been a good decade since I've done this stuff, but I remember there only being 6 or 7 for this level of movement mechanics.
You then just plug values into formulas and combine formulas until you get an answer to the question you need. Often it's easier to work backwards from the answer, figuring out what else you need to get a specific value, then looking for a formula that can answer that component, etc.
It also helps to break a complicated thing into more manageable, smaller parts, and gluing things together at the end.
 
Last edited:
srothman said:
Looking for some clarity on some basic physics scenarios, but because I'm a muppet, I need some explanations. I ducked out of high-school physics the moment I could, and I kind of regret it now...

Vertical Projectile Motion

A stone is thrown vertically upward at 20m.s-1 from a cliff 18m above the ground.
View attachment 426124

a) Determine the time it takes for the stone to pass the starting point in the way down.

Solution
v(i) = +20m.s^1
a = -9.8m.s^2
v(f) = -20m.s^1
Δt = ?

v(f) = v(i) + aΔt
-20 = +20 + (-9.8)Δt <<<<---- this is where the wheels come off completely....
Δt = 4.08s
Click to expand...

You know the saying "What goes up, must come down"? Well that this.
For the sake of your scenario, there's no drag and acceleration is constant at 9.8m.s^-2, okay?

So, what happens with a stone when you throw it up? It immediately starts accelerating downwards, until its velocity is zero, at which point it has reached its maximum height, and will start moving downwards.
It'll then continue to accelerate downwards at 9.8m.s^-2 until something stops it, like hitting the earth.
Because acceleration is constant, we know that the stone's velocity should be the same, but in the opposite direction, by the time it passes you on the way down. This can be proven by breaking the movement into parts.
Anyway, so if we take our starting velocity to be 20m.s^-1, then our ending velocity should be -20m.s^-1.

So if we plug that into this formula:
v(f) = v(i) + aΔt

We have final velocity = -20
We have initial velocity = 20
and we have acceleration constant at 9.8
and we have t unknown, which is what the question is asking you to solve.

Plugging in the values we get this:
-20 = +20 + (-9.8)Δt (Take note of the sign of the acceleration. It indicates direction.)

Solving for t, we move over the "+20" to the left side of the equasion:
-40 = (-9.8)Δt

Then we move over the -9.8:
-40/(-9.8) = Δt

Which gives us:
Δt = 4.08s
 
srothman said:
Looking for some clarity on some basic physics scenarios, but because I'm a muppet, I need some explanations. I ducked out of high-school physics the moment I could, and I kind of regret it now...

Vertical Projectile Motion

A stone is thrown vertically upward at 20m.s-1 from a cliff 18m above the ground.
View attachment 426124

a) Determine the time it takes for the stone to pass the starting point in the way down.

Solution
v(i) = +20m.s^1
a = -9.8m.s^2
v(f) = -20m.s^1
Δt = ?

v(f) = v(i) + aΔt
-20 = +20 + (-9.8)Δt <<<<---- this is where the wheels come off completely....
Δt = 4.08s

b) Determine the maximum height above the cliff reached by the stone.

Solution
v(i) = +20m.s^1
a = -9.8m.s^2
v(f) = 0m.s^1
Δy = ?

v(f)^2 = v(i)^2 +2aΔy
(0)^2 = (20)^2 + 2(-9.8)Δy
Δy = +20.41m

c) Determine the maximum velocity of the stone when it is 11m above the foot of the cliff.

Solution
v(i) = +20m.s^1
a = -9.8m.s^2
v(f) = ?

v(f) = v(i) + 2aΔy
v(f) = (20)^2 + 2(-9.8)(-7)
v(f) = 23.18 m.s^1

d) Where, relative the cliff, is the stone when it reaches a velocity of 15m.s?

Solution
v(i) = +20m.s^1
a = -9.8m.s^2
v(f) = + or - 15m.s^1
Δy = ?

v(f)^2 = v(i)^2 + 2aΔy
(+- 15)^2 = (20)^2 + 2(-9.8)Δy
Δy = +8.93m

This is straight out of a handbook, but I can't, for the love of me, understand how they get to the solutions.
Click to expand...

v(f) = v(i) + aΔt
-20 = +20 + (-9.8)Δt <<<<---- this is where the wheels come off completely....
Δt = 4.08s


Manipilation to get the Δt alone since they are asking for time it takes.

v(f) = v(i) + aΔt
-20 = +20 + (-9.8)Δt
-20-20=-9.8Δt
-40/-9.8=Δt
Δt = 4.08s

Hope this helps.
 
Simple equations of motion with the variables
u
v
a
t
s
where you need 3 of those variables to solve all the rest of the values and often initial velocity and or initial displacement is 0.
 
Last edited:
Messugga said:
Do you have the formulas for displacement?
I'll do one solution in a separate post, but I don't like spoonfeeding people and you've not really indicated what you don't understand.
Click to expand...

Thanks a mill!

these are the equations I have:

d = v(i) * t + ½ * a * t^2
v(f) = v(i) + a * t
v(f)^2 = v(i)^2 + 2 * a * d
d = (v(i) + v(f)) / 2 * t

I didn't know where the 4.08 came from, given those specific values, i.e. how to manipulate them to get the formula to work. I'm just not wired to do stuff like this, but that explanatiion made all the differrnce, thanks.

CrzWaco said:
Hope this helps.
Click to expand...

It helped a bucketload, thanks.
 
Last edited:
The way the solution is given to the first problem assumes quite a lot (i.e that you know that the velocity will be the same magnitude but in the opposite direction when the stone reaches the starting point on the way down). This can be solved by without making that assumption by using the other formula:
d = v(i)*t + 0.5*a*t^2
d = 0 (because it is back at the starting point)
v(i) = 20
a = -9.8
giving
0 = 20t + 0.5*(-9.8)*t^2
-4.9 t^2 + 20t = 0
which is a quadratic equation of the form: a*t^2 + b*t + c = 0
with a = -4.9, b = 20 and c = 0
with solutions: t = (-b +- sqrt(b^2 - 4ac))/2a
therefore t = 0 or t = 4.08
 
srothman said:
i.e. how to manipulate them to get the formula to work.
Click to expand...

Manipilation to get the Δt alone since they are asking for time it takes.

This often confusing to people because they don't know how to move things across and algebraic formula.
Let me see if I can explain.

We can agree that 1 = 1 or a = a, simple logic.
The next step is that you need to zero sum the one side of the equation to move it to the other side. You do this by subtracting an object form itself.

a+c = a+b
a+c-b = a+b-b
therefore
a+c-b=a

Similarly below:

v(f) = v(i) + aΔt
-20(-20) = +20(-20) + (-9.8)Δt

Resolves to:
-40=(-9.8)Δt

To Move the -9.8 over:
(-40)/(-9.8)=(-9.8)Δt/(-9.8)

Resolves to:
-40/-9.8=(1)Δt

(1)Δt is 1 * Δt which is Δt, so you can drop the (1)

Because a = b and b = a, we can simply swap them around
Δt = 4.08
 
Last edited:
To complicate matters further, ALWAYS keep all symbols.
v(i) = +20m.s^1
a = -9.8m.s^2
v(f) = -20m.s^1
Δt = ?

Corrected with symbols:
v(f) = v(i) + aΔt
-20m.s^1(-20m.s^1) = 20m.s^1(-20m.s^1) + (-9.8m.s^2)Δt

Resolves to: - Notice how addition and subtraction have no impact on symbol values
-40m.s^1=(-9.8m.s^2)Δt

To Move the -9.8 over:
(-40m.s^1)/(-9.8m.s^2)=(-9.8m.s^2)Δt/(-9.8m.s^2)

Resolves to: Notice how multiplication and division do have an impact on symbols
-40m.s^1/-9.8m.s^2=(1)Δt

(1)Δt is 1 * Δt which is Δt, so you can drop the (1)

In division a^x / a^y = a^(x-y)thereofore:
m.s^1 / m.s^2 = m^(1-1).s(1-2)
= m^0.s^1
=1.s
=s

Therefore symbolically correct
Δt = 4.08s
 
I'm going to have to start paying you lot tutor fees for all this help. Thanks a lot!
 
DMNknight said:
To complicate matters further, ALWAYS keep all symbols.
v(i) = +20m.s^1
a = -9.8m.s^2
v(f) = -20m.s^1
Δt = ?

Corrected with symbols:
v(f) = v(i) + aΔt
-20m.s^1(-20m.s^1) = 20m.s^1(-20m.s^1) + (-9.8m.s^2)Δt

Resolves to: - Notice how addition and subtraction have no impact on symbol values
-40m.s^1=(-9.8m.s^2)Δt

To Move the -9.8 over:
(-40m.s^1)/(-9.8m.s^2)=(-9.8m.s^2)Δt/(-9.8m.s^2)

Resolves to: Notice how multiplication and division do have an impact on symbols
-40m.s^1/-9.8m.s^2=(1)Δt

(1)Δt is 1 * Δt which is Δt, so you can drop the (1)

In division a^x / a^y = a^(x-y)thereofore:
m.s^1 / m.s^2 = m^(1-1).s(1-2)
= m^0.s^1
=1.s
=s

Therefore symbolically correct
Δt = 4.08s
Click to expand...

You lost a lot of "-"'s in the exponents. Should be m.s^(-1) and m.s^(-2) for velocity and acceleration respectively. Given how you wrote it down your result is 4.08 nothings per second rather than 4.08 seconds.
 
You must log in or register to reply here.
Back
Top
Sign up to the MyBroadband newsletter
X