A sly 1.5-kg monkey and a jungle veterinarian with a blow-gun loaded with a tranquilizer dart are 25 m above the ground in trees 70 m apart. Just as the veterinarian shoots horizontally at the monkey, the monkey drops from the tree in a vain attempt to escape being hit. What must the minimum muzzle velocity of the dart be for the dart to hit the monkey before the monkey reaches the ground?

As both the monkey and the darts are subjected to constant gravitational acceleration g = 9.8 m/s2 and both start from rest (vertically speaking). Their vertical position will always be the same. For the dart to hit the monkey, its horizontal position must be the same as the monkey's, which is unchanged before reaching the ground. Therefore, the time it takes for the dart to travel across 70 m must be less than the time it takes for the monkey to drop 25m to the ground. We can find it out using the following equation of motion

[tex]s_m = gt_m^2/2[/tex]

[tex]25 = 9.8t_m^2/2[/tex]

[tex]t_m^2 = 50/9.8 = 5.1[/tex]

[tex]t_m = \sqrt{5.1} = 2.26 s[/tex]

For the dart to takes less that 2.26 s to travel 70m, its horizontal speed must at least be 70 / 2.26 = 31 m/s