Projectile Motion Dart and Target: Proving Contact with Specific Variables

[polak333]

Homework Statement

Question: At the instant a dart is launched at a high velocity, a target (often a cardboard monkey) drops from a suspended position downrange from the launching device. Show that if the dart is aimed directly at the target, it will always strike the falling target. (use a specific set of numbers).

Variables I decided to use:
velocity (v₁= 100m/s
angle shot at = 45 degrees
time = 10 seconds

Homework Equations

d(t) = (v₁cosx)t + (v₁sinx)t - 1/2(g)(t)²
Therefore:
Horizontal - x = (v₁cosx)t)
Vertical - y = (v₁sinx)t - 1/2(g)(t)²

The Attempt at a Solution

This is the start in the problem. I'm not exactly sure how I'm supposed to prove it. I started with finding the horizontal and vertical.
Hor:
x = (v₁cosx)t)
x = (100cos45)10
x = 707m

Ver:
y = (v₁sinx)t - 1/2(g)(t)²
y = (100sin45)10 - 1/2(9.8)(10)²
y = 217m

Finding distance traveled by dart:
a² + b² = c²
707² + 217² = c²
c = 739.5m

Finding distance traveled by cardboard monkey in 10 seconds:
d = v/t
d = 9.8 * 10
d = 98m

Therefore, I know how far the monkey falls in 10 seconds and how far and high the dart ends up, but how do I prove that the dart and monkey come in contact.
Unless of course I did something with the steps above and have everything wrong.
Any help is appreciated.
Thank you.

 

[kuruman]

You have a good start. There are many ways you can answer this question, but since you are asked to invent some numbers, here is what I would do. I would choose the horizontal and vertical distance of the monkey to be something like lecture hall size, no larger than 20 m. If these are equal, then the aiming angle is 45^o, if they are not equal you can easily do the trig and find the angle. I would choose a reasonable dart speed knowing that 20 m/s is 45 miles per hour.

Then I would write two sets of kinematic equations, in the x and y directions, one set for the dart's motion and one for the monkey's motion. I would use the horizontal equation to find how long it takes the dart to get across (time of flight). Finally, I would plug the time of flight in each of the vertical motion equations separately and show that the dart and the monkey would be at the same height above ground.

 

[polak333]

Ok, so I tried it like this, this time.
/|
/ |
20m/s / |
/ | 10 m
/ |
/ |
/__45*______|
10 m

Therefore, the dart (initial velocity) is 20m/s.
The distance from the launcher is 10m.
The distance from the ground to the target hanging is 10m.
Thus the angle is 45 degrees.

Solving:
Time
t = d/t
t = 10m / 20m/s
t = 0.5 s

Dart Kinematics
X-dir
x = (20cos45*)
x = 14.14

Y-dir
y = (20sin45*) - 9.8 (I use -9.8 because up is positive, and I subtract gravity).
y = 4.43

Monkey Kinematics
X-dir
x = 0

Y-dir
y = -9.8

Finding distance monkey fell
= -9.8m/s
= -9.8m/s (0.5s)
= -4.9m
Total distance is 10m, therefore 10m - 4.9m = 5.1m above ground.

But now, wouldn't the distance for the dark before 4.34m? Isn't that what I found in the y-direction for the dart? I'm not sure what to do from here on. Should I use pythagorean theorem on the dart or is that irrelevant for this?
Thanks

 

[kuruman]

Your kinematics equations leave something to be desired. Starting from what equation did you get for the monkey

y = -9.8 ?

This says that the monkey's position above ground is -9.8 m and remains that for all time. Surely you don't mean that!

For the dart, you say
x = (20cos45*)

This also says that the horizontal position of the dart doesn't change either. Same with the y position of the dart.

You need to introduce time t as a parameter. Look up the kinematics equations and use them correctly.

 

[polak333]

I hope this is the last post.

Solving:
Time
t = d/t
t = 10m / 20m/s
t = 0.5 s

Dart Kinematics
X-dir
x = (20cos45*)t
x = (14.14)(0.5)
x = 7.07m/s

Y-dir
y = (20sin45*)t - 1/2gt²
y = 7.07 - 1/2(9.8)(0.5)²
y = 5.84m/s

Monkey Kinematics
X-dir
x = 0

Y-dir
y = -9.8t
y = -4.9m/s

So now with the time components in, I think I need to find the height of both.
This is where I still don't exactly understand what I need to do.
So I know that the y direction for the dart is 5.84m/s and the y direction for the monkey is -4.9. But I just don't know what to do with it. Do I find the distance? If yes, then how? By multiplying it by 0.5s?
Thanks.

 

[kuruman]

The monkey kinematics equations are incorrect. If you decide on using an origin for the dart, you must use the same origin for the monkey. Your monkey x equation says x = 0 and y = -9.8t. That says that at t = 0, the monkey is at point (0,0), right where the dart is. That's not the case. Initially, the monkey is at (10,10). Write kinematics equations that reflect that.

 

[polak333]

GOT IT!

Well, finally figured it all out.
Here it is for anyone who has struggled this much:

Solving:
Dart Kinematics
X-dir
Vx = (20cos45*)(0.707s)
x = 10m

Y-dir
Vy = (20sin45*)(0.707) - 1/2 (9.8)(0.707)²
y = 7.55m

Monkey Kinematics
X-dir
Vx = 0m/s
x = 10m

Y-dir
Vy = 10m - 1/2 (9.8)(0.707)²
y = 7.55m

Time hit: d/t
= 10m / 7.07 m/s
= 0.707s

THEREFORE, dart position: x = 10m, y = 7.55m
THEREFORE, target position: x = 10m, y = 7.55m
They will be at the same position and thus will hit.

 

[Delphi51]

Interesting that the general solution is no harder than the example.

 

[kuruman]

I am glad you got it, polak333.

Interesting that the general solution is no harder than the example.

Even more interesting is to note that this can be explained without equations. Both objects are in free fall, therefore their velocities change exactly the same way. Thus, their relative velocity is constant and, if the dart is initially aimed straight at the monkey, it will remain aimed straight at the monkey.