1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Free fall, hit the target, projectile problem

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    two dimension kinematics

    3. 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
  2. jcsd
  3. Sep 21, 2013 #2


    User Avatar
    Science Advisor

    The shooter is on top of another building - so your equation 1 is only correct if h is the difference between the heights of the buildings.

    Fixing that, then #2 is an equation describing the vertical motion of the dart and monkey. Essentially, it tells you when the monkey drops below the dart.

    On the other hand, #3 is talking about horizontal motion of the dart. It tells you when the dart passes the (vertically falling) monkey.

    If the monkey drops past the dart at the same time as the dart passes the monkey, they hit.

    This probably isn't right, because a dart doesn't follow a purely ballistic trajectory - it is also an aerodynamic body.
  4. Sep 21, 2013 #3
    Ok so, the question is if the dart will hit the monkey or not, so....

    55 sin (10) x t = H
    55 sin (10) x t = 79.3 - 65.2
    55 sin (10) x t = 14.1
    t = 14.1/ 55sin10

    t = 1.476

    so if they were to hit, it should hit at 1.476

    v sin(θ)t - ½gt2 = h - ½gt2

    55 (sin10) x 1.476 - (1/2)(9.8)(1.476^2) = 14.1 - (1/2)(9.8)(1.476^2)

    is that right? if it equals it?
  5. Sep 22, 2013 #4


    User Avatar
    Science Advisor

    Not quite. I agree with your calculation of t (at least as far as t=14.1/(55 sin(10)) - I haven't checked that you've put the numbers into the calculator correctly).

    That tells you that the dart and the monkey are at the same height 1.476 after the monkey jumps and the man shoots. It doesn't tell you if they are at the same horizontal position or not.

    How can you work out when they are at the same horizontal position?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted