Find the value of ##T## and distance of particle in the first ##4## seconds

[chwala]

Homework Statement

See attached. (question with solution)

Relevant Equations

Mechanics

solution is here;

I just need to understand this part $14.7 = -14.7 =9.8T$... why initial velocity upwards is a negative value? or i am interpreting it wrongly.

...........
In my reasoning,

$v=u+at$
$0=14.7 + (-9.8)t$
$t_1=1.5$

in reverse direction, from top to start point $T$,

$14.7=0+9.8t$
$t_2=1.5$

$T=1.5+1.5=3$seconds

For second part i have the equation,

particle moving up vertically,

$v^2=u^2+2as$

$0 = 14.7^2 + (2× -9.8s)$
$216.09=19.6s$
$s=11.025$m

and for particle moving downwards,

$v=u+at$
$v=0 + 9.8× 2.5$
$v=24.5$

$24.5^2=0+19.6s$

$s=30.625$m

thus $s_{total} = 11.025+30.625=41.65≅41.7$m

 

[Herman Trivilino]

I just need to understand this part 14.7=−14.7=9.8T... why initial velocity upwards is a negative value?

It's $14.7=-14.7+9.8T$.
Assuming these are values substituted into $v=v_o+at$ we have $v=+14.7$, $v_o=-14.7$, and $a=+9.8$. Evidently the author has chosen the downward direction to be positive, thus the initial upward velocity is negative.

 

[Orodruin]

… and physics does not care which direction you consider positive. The author chose down as the positive direction, you chose up. Both conventions give the same result as long as you are consistent.