Find the velocity of the stone just before it hits the ground

[chwala]

Homework Statement

Highlighted question in Red.
Question.5

Relevant Equations

Mechanics

In my lines i have,

$h(t)=ut+\dfrac{1}{2}at^2$ where $u=0$,

with $a=10$,

$h = \dfrac{1}{2}×10t^2$ ....................1

...also,

$h-14.7 = \dfrac{1}{2}×10 (1.4-t)^2$ ..................2

1- 2 gives,

$14.7 = 14t -9.8$

$24.5=14t$

$t=1.75$

Therefore, $v = 10× 1.75=17.5$ m/s , with $u=0$.

correct? Cheers

 

[PeroK]

That solution is incomprehensible to me.

 

[nasu]

If by "u" you mean initial speed, it is not zero. The stone is "thrown downwards".

 

[kuruman]

What is $h$ in the equation you wrote $~h-14.7 = \dfrac{1}{2}×10 (1.4-t)^2~?$
Hmm $\dots$ Let's see.
At $~t=0$, $h=(14.7+\dfrac{1}{2}\times 10\times 1.4)~\text{m}=21.7~\text{m}.$
At $~t=1.4~\text{s}$, $h=14.7~\text{m}.$
This doesn't match what is given.

 

[WWGD]

I agree with @nasu . Initial velocity would be $0$ if the stone had been dropped or had somehow fallen.

 

[Lnewqban]

.

 

[PeroK]

The calculated velocity can’t be correct ..

I think it is!

 

[nasu]

It's the final velocity who is 17.5 m/s and not the initial. The question asks for the final velocity (when it hits the ground). vo is about 3.5 m/s.

 

[kuruman]

Oops. Sorry about that. I fixed in my mind that it was the initial velocity. I deleted the post to avoid unnecessary confusion.

 

[PeroK]

Let's test if it is by writing a more conventional SUVAT equation and substituting the given numbers and the purported solution.

The height of the stone above ground at any time $t$ is
$h=h_0+v_0t-\frac{1}{2}gt^2$
With $h_0=14.7~$m, $g=10~\text{m/s}^2$ and an assumed $v_0=-17.5~$m/s, the height above ground at specific time $t=1.4~$s is predicted to be
$h=(14.7-15.7\times1.4-\frac{1}{2}\times 10\times 1.4^2)~\text{m}=-29~\text{m}.$
That is 29 m below ground.

That's equally incomprehensible to me.

The average velocity is $10.5m/s$, which would be at $t = 0.7s$. So, initial velocity is $3.5m/s$ and final velocity $17.5 m/s$.

All downwards.

 

[kuruman]

That's equally incomprehensible to me.

The average velocity is $10.5m/s, which would be at$t = 0.7s$. So, initial velocity is$3.5m/s$and final velocity$17.5 m/s##.

See post #9.

 

[kuruman]

Just to redeem myself, another way to approach this problem is to use the SUVAT equation with the final velocity replacing the initial velocity, $$h=h_0+v_f~t-\frac{1}{2}at^2.$$Of course here $a=-g$.

 

[Lnewqban]

It's the final velocity who is 17.5 m/s and not the initial. The question asks for the final velocity (when it hits the ground). vo is about 3.5 m/s.

Thank you both, @nasu and @PeroK
Deleted post with incorrectly assumed velocity.

 

[chwala]

In my reasoning, in reference to my initial velocity being $0$, the person throwing the stone had the clock ticking for some time (stone still on his hands),that is for$1.75-1.4$ seconds before throwing ball ... that landed $1.4$ seconds later on the ground... that was my reasoning and it looks like it isn't correct.

 

[chwala]

I agree with @nasu . Initial velocity would be $0$ if the stone had been dropped or had somehow fallen.

You see 'dropped and thrown' ...to me I have to try and understand the language. I think now I see the two aren't the same.

 

[chwala]

It's the final velocity who is 17.5 m/s and not the initial. The question asks for the final velocity (when it hits the ground). vo is about 3.5 m/s.

Boss I was wondering how you got the $u = 3.5$m/s.

arrrggh I see that...


From
...
$14.7 = 1.4u +9.8$
Which means,

$v^2= 3.5^2 + (2 •10•14.7)$

$v^2 = 306.15$

$v= 17.5$


Cheers man!

 

[haruspex]

another way to approach this problem is to use the SUVAT equation with the final velocity replacing the initial velocity,

Indeed.
The way I describe SUVAT is that there are five standard variables, s, u, v, a, t (hence the name), where u and v are initial and final velocities.
Correspondingly, there are five standard equations, each omitting one of the variables. So the first step is to identify which three you know the values of and which one you want to find, then pick the equation using those four.
It gets trickier when connected with other movements. E.g. you might know s and u only and want to find a. If the same time is involved in another process then the four relevant variables are s, u, a, t.