Respuesta :
Explanation:
Given
initial velocity(v_0)=1.72 m/s
[tex]\theta =44.8{\circ}[/tex]
using [tex]v^2-u^2=2as[/tex]
Where v=final velocity (Here v=0)
u=initial velocity(1.72 m/s)
a=acceleration [tex](gsin\theta )[/tex]
s=distance traveled
[tex]0-(1.72)^2=2(-9.81\times sin(44.8))s[/tex]
s=0.214 m
(b)time taken to travel 0.214 m
v=u+at
[tex]0=1.72-gsin(44.8)\times t[/tex]
[tex]t=\frac{1.72}{9.8\times sin(44.8)}[/tex]
t=0.249 s
(c)Speed of the block at bottom
[tex]v^2-u^2=2as[/tex]
Here u=0 as it started coming downward
[tex]v^2=2\times gsin(44.8)\times 0.214[/tex]
[tex]v=\sqrt{2.985}[/tex]
v=1.72 m/s
