Free fall, hit the target, projectile problem

  • Thread starter annakwon
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  • #1
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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.
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
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Ignore the setup for a moment.
Neglecting air friction - which falls faster, the dart or the monkey?
Ergo - where do you have to be aiming to hit the monkey?
Which suspect is doing that?

Of course, I have not sketched the situations so it may not be the problem I'm thinking of.
Anyway - your reasoning looks OK. You are finding the time that the dart from each suspect has the same height as the monkey and then check that the x-positions also coincide at that time.
 

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