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 mhow much additional time will pass before the ball passes the tree branch on the way back down?

The above problem is of the free-falling body type.

The ball is thrown upwards with the initial velocity of 15.0 m/s.

The height covered by the ball is 7.00m.

Let’s understand the question by creating a schematic diagram as done below.

As mentioned in the figure, part A represents the point mentioned as A.

Here the velocity is needed, and the time is also not given.

U=15.0 m/s

g= -9.8 m/s²

v=?

d= 7.00 m

T=?

The velocity of the rock is

\(\begin{array}{c}{v^2} - {u^2} = 2gd\\{\left( v \right)^2} - {\left( {15} \right)^2} = 2\left( { - 9.81} \right)\left( 7 \right)\\v = \sqrt {87.66} \,\,m/s\\V = 9.36\,m/s\end{array}\)

Hence the final velocity in case of part A is \(9.36 m/s\).

Here the stone is travelling from one point to another; hence the final velocity for part A will become the initial velocity for part B.

Given data:

U=9.36 m/s

g=-9.81 m/s²

V= -9.36 m/s

T=?

Hence the time can be calculated using the equation below.

\(\begin{array}{c}v = u + gt\\ - 9.36 = 9.36 + \left( { - 9.81} \right)t\\t = \frac{{ - 18.72}}{{ - 9.81}}\\t = 1.908\,s\end{array}\)

Hence the additional time before the ball passes the tree branch on the way back down is \(1.908 s.\)