Free fall velocity and distance

[poh]

Homework Statement

A 0.1 kg nut is thrown by a squirrel off of the top of a 10 m tall tree. It is thrown with an initial upward velocity of 4 m/s and an initial horizontal velocity of 3 m/s.
a) How long till it reaches the top of its arc?
b) From the top of its arc how long till it reaches the ground?
c) How far will it travel horizontally from the moment it is let go to when it reaches the ground? (Consider the combination of the time in both parts a and b.)

Homework Equations

Vf=Vi + at
Xf= Xi + Vit +1/2at^2

The Attempt at a Solution

am I applying information correctly?
to find time i should Yf= Yi+at ?
I am lost for a) I should use Vf= Vi + at
and for B) Yf= Yi + Viyt + 1/2at^2

 

[haruspex]

to find time i should Yf= Yi+at ?

Did you mean that? I assume Yi and Yf are heights.

Xf= Xi + Vit +1/2at^2

If X means horizontal, what is a?

for a) I should use Vf= Vi + at

Yes.

for B) Yf= Yi + Viyt + 1/2at^2

You can, but define exactly what you mean by Yi, Vi and t here.

 

[poh]

Did you mean that? I assume Yi and Yf are heights.

Yi is initial height
Yf is final height

If X means horizontal, what is a?

a=0 right?

You can, but define exactly what you mean by Yi and t here.

Yi - initial height and t is the time

 

[haruspex]

Yi is initial height
Yf is final height

Then the equation is wrong. The right hand side, at, has dimension of velocity. Elsewhere you have correctly used Vf= Vi + at.

a=0 right?

In that equation, yes.

Yi - initial height and t is the time

I wrote "exactly". You are asked to consider the trajectory of the nut in two stages, from being thrown to reaching top of arc, then from top of arc to ground. Which point are you considering "initial" in this equation and to what interval does t refer?

 

[poh]

for b) i tried Yf= yi + Viyt + 1/2 at^2
0=10m +4m/s*t + 1/2(9.8m/s)t^2
t= 2*(10m+4m/s)/9.8 m/s
t= 28/9.8s
t= 2.9s
t is time to reaches the ground from the top
Or i should assume my Yi is 0?

 

[haruspex]

0=10m +4m/s*t + 1/2(9.8m/s)t^2

t is time to reaches the ground from the top

Careful with signs.
Let me decode your equation, on the assumption that up is positive:
For a certain motion lasting time t, the final height is zero, the initial height is 10m, the initial speed is 4m/s upwards, and the acceleration is 9.8m/s² upwards.
Anything in your post you'd like to correct?

 

[poh]

Careful with signs.
Let me decode your equation, on the assumption that up is positive:
For a certain motion lasting time t, the final height is zero, the initial height is 10m, the initial speed is 4m/s upwards, and the acceleration is 9.8m/s2 upwards.
Anything in your post you'd like to correct?

acceleration should be -9.8m/s^2

 

[haruspex]

acceleration should be -9.8m/s^2

Right, and what about the meaning of t in your equation?

t is time ... from the top

For a certain motion lasting time t ... the initial height is 10m

 

[poh]

Right, and what about the meaning of t in your equation?

t is time how much time takes to fall ?

 

[haruspex]

t is time how much time takes to fall ?

From where?

 

[poh]

From where?

from 10 m tall tree to the ground

 

[haruspex]

from 10 m tall tree to the ground

Yes. So what do you get for that time?

 

[poh]

Yes. So what do you get for that time?

I got t= 2.9s

 

[haruspex]

I got t= 2.9s

There's another problem I had not previously noticed, that your next line was:

t= 2*(10m+4m/s)/9.8 m/s

This is completely wrong. You cannot add a distance to a speed. Your equation has a 't' term and a 't²' term. How do you solve equations like that?

 

[poh]

This is completely wrong. You cannot add a distance to a speed. Your equation has a 't' term and a 't2' term. How do you solve equations like that?

divede t by itself or I should use square roof to take out t^2 ?When I assume Viy = 0 then I got 1.4s

 

[haruspex]

divede t by itself or I should use square roof to take out t^2 ?When I assume Viy = 0 then I got 1.4s

It is a quadratic equation. Use the formula.

 

[poh]

It is a quadratic equation. Use the formula.

i got 1.1s

 

[haruspex]

i got 1.1s

I get something different. Please post your working.

 

[Nik_2213]

You have excluded air-friction etc, should mention that you've done so to show 'Due Care'...

 

[poh]

I get something different. Please post your working.

I did mistake in calculations.. my new calculations are 1.8 or 1.1

 

[haruspex]

I did mistake in calculations.. my new calculations are 1.8 or 1.1

Again, be careful with signs.
You also need a bit more precision.

 

[poh]

Again, be careful with signs.
You also need a bit more precision.

-1.1 and 1.9s

 

[haruspex]

-1.1 and 1.9s

Yes, that's better.
Clearly you want the pisitive answer, but the negative one also has a physical meaning. Can you see what it is?
Oh, and remember that what you have found is the time from being thrown up. Is that what you were asked for?

 

[poh]

Yes, that's better.
Clearly you want the pisitive answer, but the negative one also has a physical meaning. Can you see what it is?

Negative means that's going down?

 

[haruspex]

Negative means that's going down?

Sort of.
The equation you solved is valid forwards and backwards in time. If you imagine the nut being thrown from ground level 1.1s earlier, with just the right speed and angle, it would have passed the monkey at "time 0" and continued on exactly the same trajectory described by your equation.

And please note the edit to my preceding post.

 

[poh]

Sort of.
The equation you solved is valid forwards and backwards in time. If you imagine the nut being thrown from ground level 1.1s earlier, with just the right speed and angle, it would have passed the monkey at "time 0" and continued on exactly the same trajectory described by your equation.

And please note the edit to my preceding post.

I was asked how longs takes to reach the ground. so answer is 1,9s

How long till it reaches the top of its arc?
is Vf=Vi + at where a=0 and t=1.9s ?

 

[haruspex]

I was asked how longs takes to reach the ground. so answer is 1,9s

How long till it reaches the top of its arc?
is Vf=Vi + at where a=0 and t=1.9s ?

a=0 is for horizontal motion. For a) and b) we only need consider vertical motion.
The questions were
a) How long till it reaches the top of its arc?
b) From the top of its arc how long till it reaches the ground?
In post #1 you had the right equation for a), but I did not see what answer you got. It should not be 1.9s.
In post #22 you got 1.9s, but that was the time from where to where?

 

[poh]

a=0 is for horizontal motion. For a) and b) we only need consider vertical motion.
The questions were
a) How long till it reaches the top of its arc?
b) From the top of its arc how long till it reaches the ground?
In post #1 you had the right equation for a), but I did not see what answer you got. It should not be 1.9s.
In post #22 you got 1.9s, but that was the time from where to where?

ah I see now so Vf=Vi+at where Vf=0 Vi 4m/s and a=9.8 m/s t=? so
t= (Vf-Vi)/a t= -4m/s / -9.8 m/s^2
t= .41s

In post #22 you got 1.9s, but that was the time from where to where?

from top to the ground or I should 1.9 s- 1.1s = .8s and that's time from top to ground ?

 

[haruspex]

t= .41s

Yes, that is the time for part a).

from top to the ground

What was the value of V_i in the equation you used to get the 1.9s? At what point of the trajectory was that the nut's vertical velocity?

 

[poh]

What was the value of Vi in the equation you used to get the 1.9s? At what point of the trajectory was that the nut's vertical velocity?

Vi = 4m/s I think it was all the time vertical velocity

 

[haruspex]

Vi = 4m/s I think it was all the time vertical velocity

Yes, it was 4m/s, but it only had that vertical velocity at one moment, the moment when it was thrown.
So the 1.9s was from the throw to landing on the ground. You are asked for the time from the top of the arc to landing on the ground.

 

[poh]

Yes, it was 4m/s, but it only had that vertical velocity at one moment, the moment when it was thrown.
So the 1.9s was from the throw to landing on the ground. You are asked for the time from the top of the arc to landing on the ground.

it's take same amount of time to go up as go to down. -1.1s is from ground level so when 1.9s-1.1s its where they meet and point is 0 and the diffrence is time from the top of the arc to ground,

 

[haruspex]

it's take same amount of time to go up as go to down. -1.1s is from ground level so when 1.9s-1.1s its where they meet and point is 0 and the diffrence is time from the top of the arc to ground,

I have read that several times and still cannot decipher it.
There are three points in time:

1. Nut is thrown; height 10m, vertical speed +4m/s
2. Nut reaches highest point; vertical speed 0
3. Nut hits ground; height 0

You found that the time from 1 to 3 is 1.9s, and the time from 1 to 2 is 0.41s. In b), you are asked for the time from 2 to 3.

 

[poh]

I have read that several times and still cannot decipher it.
There are three points in time:

1. Nut is thrown; height 10m, vertical speed +4m/s
2. Nut reaches highest point; vertical speed 0
3. Nut hits ground; height 0

You found that the time from 1 to 3 is 1.9s, and the time from 1 to 2 is 0.41s. In b), you are asked for the time from 2 to 3.

I see now Thank you :D

 

[neilparker62]

Alternate method using average velocity with upwards positive so g = -9.8 ms^-2 and ground level (h) = -10m from projectile launch position. Doesn't add much to what's been discussed above so perhaps more of a summary.

Initial vertical velocity: $$v_i=4\;ms^{-1}$$
Final vertical velocity: $$v_f=-\sqrt{{v_i}^2+2gh}\; ms^{-1}$$
Average vertical velocity: $$ v_{av}=\frac{v_i+v_f}{2}\; ms^{-1}$$
Time of travel: $$ t = \frac{h} {v_{av}} \;s $$
Horizontal displacement: $$ 3t \;m$$

Hence:

a) -4/g
b) t - (-4/g)
c) 3t