You throw a ball straight up with an initial velocity of 15.0 m/s. it passes a tree branch on the way up at a height of 7.00 m. how much additional time will pass before the ball passes the tree branch on the way back down? 50. a kangaroo can jump over an object

Additional time of 1.91 seconds is required before the ball passes the tree branch on the way back down.

Step-by-step explanation:

Consider upward direction as positive,

Here we need to find time interval between first and second 7 m displacements.

We have equation of motion, s = ut + 0.5 at²

         Displacement, s = 7 m

         Initial velocity, u = 15 m/s

         Acceleration, a = -9.81 m/s²

Substituting

       s = ut + 0.5 at²

       7 = 15 t + 0.5 x -9.81 x t²

       4.905 t² - 15 t + 7 = 0

    So at t = 0.57 s and t = 2.48 s the displacement is 7 m

    Interval between them = 2.48 - 0.57 = 1.91 s

 So additional time of 1.91 seconds is required before the ball passes the tree branch on the way back down.

15.36 per mile

but forgot to round

i am pretty sure

my computer is slow i can not see it sorry