Answer:
5 sec
Explanation:
t1 is time moving upwards and t2 is time moving downwards.
For v=0 when the ball is highest
s= -1/2(-g)(t1)^2
where s=AB=19.6m and g=9.8m/s^2
Solving t1=2 sec
The height at t1 is 44.1m
and the ball would move through gravity with zero initial velocity.
s= 1/2g(t2)^2
where s=44.1m
Solving t2=3sec
Total time consumed is T= t1+t2= 5sec
The value of time needed to complete the upward and the downward motion is, T=5sec .T is the time at which the ball first hits the ground.
The time period is found as the ratio of the displacement and the velocity. Its unit is second and denoted by t
The given data in the problem is;
The time required to move upwards is,t₁
The time required to move downwards is,t₂
The distance traveled is,s= 19.6m
The gravitational acceleration is, g=9.81 m/s
At the highest position, the velocity of the ball is zero;
The time required to move upwards is found as;
[tex]\rm S_1 = ut + \frac{1}{2} at^2 \\\\ \rm S_1 = \frac{1}{2} (-g)t_1^2 \\\\ t_1= \sqrt{\frac{2S_1}{g} } \\\\ t_1= \sqrt{\frac{2 \times 19.6 }{9.81} } \\\\ t_1= 2 \ sec[/tex]
The time required to move downward is found as;
[tex]\rm S_2 = ut + \frac{1}{2} at_2^2 \\\\ \rm S_1 = \frac{1}{2} (-g)t_2^2 \\\\ t_2= \sqrt{\frac{2S_2}{g} } \\\\ t_2= \sqrt{\frac{2 \times 44.1 }{9.81} } \\\\ t_2= 3 \ sec[/tex]
The total time period is the sum of the time required to move upwards and the time required to move downwards.
T =t₁+t₂
T=2+3
T=5 sec
Hence the value of T will be 5 sec.
To learn more about the time period refer to the link;
https://brainly.com/question/569003
