Abasketball is thrown up into the air. it is released with an initial velocity of 8.5 m/s. how long does it take to get to the top of its motion?

Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 8.5 m/s

     Final velocity, v = 0 m/s    - At maximum height

     Time, t = ?

     Acceleration , a = -9.81 m/s²

     Substituting

                      v = u + at  

                      0 = 8.5 + -9.81 x t

                      t = 0.867 s

  It takes 0.867 seconds to get to the top of its motion

0.87 s

Explanation:

initial velocity, u = 8.5 m/s

Let it takes time t to reach to maximum height. At maximum height the velocity is zero, so, v = 0

Use first equation of motion

v = u - gt

where, g be the acceleration due to gravity

0 = 8.5 - 9.8 t

t = 0.87 s

Thus, the time taken to reach at top is 0.87 s.