Answer:
Explanation:
Given
height from which ball is dropped [tex]h_1=9.5\ m[/tex]
and it rises to a height of [tex]h_2=5.7\ m[/tex]
A person caught the ball in mid-way i.e. at a height of [tex]h_0=1.2\ m[/tex] from pavement
time taken to reach the bottom is given by
[tex]y=ut+\frac{1}{2}at^2[/tex]
where y=displacement
u=initial velocity
a=acceleration
t=time
here initial velocity is zero i.e.
[tex]h_1=0+\frac{1}{2}gt_1^2[/tex]
[tex]t_1=\sqrt{\frac{2h_1}{g}}[/tex]
Similarly to reach of height [tex]h_2[/tex] time taken is [tex]t_2[/tex]
[tex]t_2=\sqrt{\frac{2h_2}{g}}[/tex]
The ball is caught in mid-way i.e. ball travel a distance of [tex]h_3=5.7-1.2=4.5\ m[/tex]
time taken is [tex]t_3=\sqrt{\frac{2h_3}{g}}[/tex]
total time taken [tex]t=t_1+t_2+t_3[/tex]
[tex]t=\sqrt{\frac{2h_1}{g}}+\sqrt{\frac{2h_2}{g}}+\sqrt{\frac{2h_3}{g}}[/tex]
[tex]t=\sqrt{\frac{2\cdot 9.5}{9.8}}+\sqrt{\frac{2\cdot 5.7}{9.8}}+\sqrt{\frac{2\cdot 4.5}{9.8}}[/tex]
[tex]t=3.429\ s[/tex]
total time =3.429 s
