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

Short answer

The additional time before the ball passes the tree branch on the way back down is 1.908 s

Step-by-step solution

Making schematic diagram of and understanding the situation

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.

Final velocity in part A

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

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

Hence the final velocity in case of part A is $9.36m/s$.

Final time covered in part B

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.

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

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