A monkey is free falling at a building at height 79.3 Meters. At the same time a dude shoots a dart from a building 65.2 meters tall, 81.0 meters away from monkey building, 55 m/s speed, angle at 10 degree.
two dimension kinematics
The Attempt at a Solution
what i think....
1. In order for the dart to hit the monkey, they must intersect vertically directly under the monkey's starting position at some point above the ground.
2. Setting the vertical positions of the monkey and the dart equal to each other yields the following equation:
#1: v sin(θ)t - ½gt2 = h - ½gt2
which simplies to....
#2: v sin(θ)t = h
Then other thing i can use is... #3 v cos(θ)t = Distance of X
which will give me time.
I'm a bit confused from here. So the question needs "time" since i need to know what time the dart will hit the monkey (as monkey is falling down). I'm having hard time what i need to put in for the height. since its from building to building, not building to zero, do i subtract the two?
if my logic seems flawed, please help me