Question

Let's say that the dart is launched with initial velocity components of 32.0 m/s horizontally and 10.88 m/s vertically. Imagine that this is taking place on a planet where the acceleration due to gravity is g = 9.00 m/s^2 2 , and neglect air resistance. What is the distance between the dart and the monkey at the instant the dart passes directly below the monkey? Note that the simulation shows that the monkey starts 40 m higher, and 100 m horizontally, from where the dart is launched, and you can make use of those numbers in your calculation. _______ m

Answer 1

Answer:

the distance that the dice is under the monkey 6 m

Explanation:

This is a projectile throwing exercise for the dart and free fall for the monkey, as we are asked for the distance when the dart passes under the monkey, so by that time the dice is 100 m away from the monkey, let's calculate how long it takes for the dart to reach this point

          X = Vox t

          t = X / Vox

          t = 100/32

          t = 3,125 s

With this time let's calculate the height of the dart and height that the monkey has descended

Dart

        Y = Voy t - ½ g t²

       Y = 10.88 3.125 - ½ 9 3125² = 34 - 43.94

       Y = - 9.94 m

Monkey

Let's calculate how much descended in this time, see that the monkey is 40 m higher than the dart

        Ym = Yo + Vot - ½ g t²

We assume that the monkey is released so that its Vo is zero

        Ym = Yo - ½ g t²

        Ym = 40 - ½ 9 3,125² = 40 -43.94

        Ym = -3.94 m

We already have the distances of each one, the difference between them is the subtraction of the two

         ΔDY = Yd -Ym

         ΔY = 9.94 -3.94

         ΔY = 6 m

This is the distance that the dice is under the monkey