Answer:
in both the cases the time will be same
Step-by-step explanation:
As we know that the chunk is projected horizontally
so here the initial speed of the chunk is given as
![v_x = v_o](https://img.qammunity.org/2020/formulas/physics/high-school/7bir9a7dtzrtsxux236f2ecwfxubp77zt0.png)
![v_y = 0](https://img.qammunity.org/2020/formulas/physics/middle-school/cj749qbt3niee6m7yccfmwxud8ym7gf9pa.png)
now in order to reach the ball to the ground we can say
![h = (1)/(2)gt^2](https://img.qammunity.org/2020/formulas/physics/high-school/sraezkvtcaichjbbl2ksm3rxl7tfdqe81x.png)
so we will have
![t = \sqrt{(2h)/(g)}](https://img.qammunity.org/2020/formulas/physics/middle-school/en2rbn9i95dhpa4i78s5vjuts31a8eqpkq.png)
now if another ball is just dropped from the same height then its initial speed is also zero in both direction
so again we have
![y = (1)/(2)gt^2](https://img.qammunity.org/2020/formulas/physics/middle-school/wjw8tesn3zhx3rpvshcvpfllabscpzxkqe.png)
![t = \sqrt{(2h)/(g)}](https://img.qammunity.org/2020/formulas/physics/middle-school/en2rbn9i95dhpa4i78s5vjuts31a8eqpkq.png)
so we can say that in both the cases the time will be same