- #1
annakwon
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Homework Statement
monkey is a building 79.3 meters tall. as he was free falling, he was shot by a suspect with a dart. which suspect hit the monkey?
Suspect a - on a building 65.2 tall, 81meters away from monkey (building to building horizontally), at velocity 55, at angle 10
suspect b - 21.5/40/56.2/59
suspect c - 6.5/59.3/45.2/47
Homework Equations
kinematic motion
The Attempt at a Solution
Monkey's vertical height from ground = y1 = 79.3 - 1/2gt² = 79.3 - 4.905t²
Dart's vertical height from ground = y2 = 65.2 + Voy(t) - 1/2gt²
Voy = 55(sin 10°) = 9.55 m/s
y2 = 65.2 + 9.55t - 4.905t²
The only time instant monkey can be hit by dart is when:
y1 = y2
79.3 - 4.905t² = 65.2 + 9.55t - 4.905t²
79.3 = 65.2 + 9.55t
9.55t = 14.1
t = 14.1/9.55 = 1.4764 s <= critical time instant
horizontal displacement of dart at t = R
R = Vox(t)
Vox = 55(cos 10°) = 54.2 m/s
R = (54.2)(1.4764) ≈ 80.0 m <=> 81.0 m {dart misses monkey by 1 m}
i can do the other two, but i just need to make sure if my logically thought process is in the right direction.