Since monkey is dropped from the tree as soon as He shoot the dart on it
so here the time taken by the monkey to reach the ground is given by kinematics
so it is given as
![y = (1)/(2)gt^2](https://img.qammunity.org/2020/formulas/physics/middle-school/wjw8tesn3zhx3rpvshcvpfllabscpzxkqe.png)
given that
y = 27 m
now we have
![27 = (1)/(2)(9.8)t^2](https://img.qammunity.org/2020/formulas/physics/high-school/em3jc88ht820jdt4lh5wp85c4pfvoh4kbs.png)
![t = 2.35 s](https://img.qammunity.org/2020/formulas/physics/high-school/99ezx56af5hvf6zigapy2yhqjub9wtx0lo.png)
now the dart should reach the monkey in same time
so the horizontal distance between dart and monkey is given as
![x = 98 m](https://img.qammunity.org/2020/formulas/physics/high-school/7adjlxlhmwveo7ba3q1o9pdjtuwyxtjr7n.png)
time = 2.35 s
speed = distance / time
now from above equation we have
![v = (98)/(2.35)](https://img.qammunity.org/2020/formulas/physics/high-school/zmgxexqj6ax80zawxiz9jhmzbamo7c8uow.png)
![v = 41.75 m/s](https://img.qammunity.org/2020/formulas/physics/high-school/75qehj5grvfe6732juqm62606l84nykijb.png)