Find h in the problem involving a stone thrown upwards

[chwala]

Homework Statement

See attached.

Relevant Equations

kinematics

Was the approach used by the referenced tutor correct?

In my understanding, the set up equation ought to be,
$s=-h, u=16$ as the stone is being thrown upwards and $g=-9.8$
giving me,

$-h = 64 - \dfrac{1}{2}×9.8 ×16$

$-h=-14.4$m

$h=14.4$m , or it does not really matter.

 

[Steve4Physics]

Your tutor used a sign convention where downwards is positive.
You used a sign convention where upwards is positive.
You both solved the problem correctly and got the same answer.
One approach isn't better than the other IMO.

Edit: technically it would be better to round the final answer to 2 significant figures, for consistency with the given data.

 

[chwala]

Your tutor used a sign convention where downwards is positive.
You used a sign convention where upwards is positive.
You both solved the problem correctly and got the same answer.
One approach isn't better than the other IMO.

Edit: technically it would be better to round the final answer to 2 significant figures, for consistency with the given data.

Not my tutor though ... the term 'tutor' was in reference to the attached downloaded pdf. Cheers though.

 

[chwala]

For part (b) this is clear,

but i can also use,

$v=u + at$

$v= 0 + 9.8(4-1.63265) = 23.2$ m/s

I made use of $\dfrac{ds}{dt}$, just incase the question on $1.63$ comes up.

cheers.

 

[Steve4Physics]

For part (b) this is clear,

View attachment 352465

but i can also use,

$v=u + at$

$v= 0 + 9.8(4-1.63265) = 23.2$ m/s

Or even:
$v = u + at = 16 + (-9.8) \times 4 =-23.2$ m/s
Speed = |v| = 23.2 m/s

 

[chwala]

This is a related problem, allow me to post it here rather than start a new post;

The steps are clear;

also we could solve it by simultaneous equation, that is by use of velocity and distance formula (suvat), we shall have the simultaneous,

$34=u+10a$

$480=20u+100a$

$u=14$ m/s.

cheers, great day.