Question

a hunter shoots a tranquilizer dart straight at a monkey who is hanging on a tree branch that is 1.5 meters high. the tree is 16 meters away from the hunter. the man shot the dart at the monkey with an initial sideways velocity of 40 m/s, and at the same time, the monkey let go of the branch. what time will the monkey reach the same height of the arrow? t = ___ s

Explanation:

Step1: Analyze horizontal motion

The horizontal motion of the dart is a uniform - motion. The horizontal distance $x = 16$ m and the horizontal velocity $v_x=40$ m/s. In horizontal direction, the formula is $x = v_x t$.

Step2: Solve for time

We can re - arrange the formula $x = v_x t$ to get $t=\frac{x}{v_x}$. Substitute $x = 16$ m and $v_x = 40$ m/s into the formula. So $t=\frac{16}{40}=0.4$ s. The vertical motion of the monkey (free - fall) and the vertical component of the dart's motion (projectile motion) are not needed to solve for the time when the monkey and the dart are at the same height horizontally.

Answer:

$0.4$