Free fall, hit the target, projectile problem

[annakwon]

Homework Statement

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.

Homework Equations

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

 

[Ibix]

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.

 

[annakwon]

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

so...
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?

 

[Ibix]

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?

 

[Nickie]

correct it.

First of all, your attempt at a solution is a good start. You are right in thinking that 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. This means that the horizontal distance traveled by the dart must be equal to the horizontal distance between the two buildings, which is 81.0 meters.

Next, you correctly identified the two equations that you can use to solve for the time it takes for the dart to hit the monkey. However, you need to be careful with the values you use for the variables in these equations.

For the first equation, you have correctly identified that the height of the monkey is equal to the initial height of the dart (79.3 meters). However, the initial velocity of the dart is not simply 55 m/s. This is the magnitude of the velocity, but you also need to take into account the angle at which the dart is launched. This angle will affect the horizontal and vertical components of the velocity. In order to use this equation, you will need to find the vertical component of the velocity, which can be calculated using the formula v*sin(θ), where θ is the angle of 10 degrees.

For the second equation, you have correctly identified that the horizontal distance traveled by the dart is equal to the distance between the two buildings (81.0 meters). However, again, you need to be careful with the values you use for the variables. The time you calculated from the first equation is the total time it takes for the dart to hit the ground, not just the time it takes for the dart to travel the horizontal distance. You will need to use this time value and the horizontal component of the velocity (v*cos(θ)) to solve for the distance traveled in the horizontal direction.

Once you have both equations set up correctly, you can solve for the time it takes for the dart to hit the monkey. Then, you can use this time value and the equations of motion to calculate the position of the monkey at that time, and see if it intersects with the position of the dart.

Overall, your approach is correct, but be sure to carefully consider the values you use for the variables in the equations. Also, it is always a good idea to draw a diagram to help visualize the problem and make sure you are using the correct values. Keep up the good work!