Question

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

Answer 1

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

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

`t=(-(-15)\pm √((-15)^2-4* 4.905* 7))/(2* 4.905)=(15\pm √(87.66))/(9.81)=(15\pm 9.36)/(9.81)\\\\t=2.48s \texttt{ or } t=0.57s`

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.