How to Calculate Horizontal Projectile Motion on the Moon

[mohlam12]

hello, please i need help in a problem that I don't know how to do. it s kinda complicated but yeah, just try !

here is the scene...you are on the moon (gravity = 1.624 m/ss). you have won the "chance of a lifetime to shoot a teacher with high powered dart gun" contest. you know that the high poered darts are air friction resistant (not slowed down horizontally by air) you have determined the muzzle velocity of the gun to be _____ m/sec since you found that if the gun is shot vertically on earth, the dart has 5.3 second hang time. Now back to the moon, you are standing 15 away from the target whose nose at 1.85 meters high makes an excellent target to sight on, since it is the same height as the gun. The gun is aimed horizontally! your only requirement is to determine within 5cm how far below the horizontal line of sight will you hit, if it makes it there at all.

The dart will hit __________m below the line of sight or ___________m in front of the teacher.


Please please please, i tried over and over again to get the problem and to solve it but i coulsnt, I am an exchange student and i already had difficulties to get the problem. Please try to give me hints!

I already got that the vertival muzzle velocity on Earth is 25.97 m/s
Vi= 0 - 9.8 * (5.3/2)
Vi=25.97
and on Earth the dart will go 34.41 m up (Vi.t+.5at^2 --> 25.97*(5.3/2)+.5*9.8*(5.3/2)^2 = 34.441 ) but i still don't know how to determine the muzzle velocity on the moon if we have the muzzle velocity on earth.
thanks,

 

[learningphysics]

I think you should assume the muzzle (initial) velocity is the same on the moon.

 

[mohlam12]

ok, that s what i was thinkin. but, how can i know how much the dart will hi below the line of sight on the moon. we never learned anythin about line of sight, and i didn t even get the problem, if someone at least can explain it in different and simple way
thx

 

[HallsofIvy]

You are making this much too difficult! You have two equations- one for horizontal motion, the other for vertical. Take the initial (horizontal) velocity to be what you calculated. Find the time a dart with that speed takes to go the 15 (meters?) to the target (you are given that there is no friction and gravity only acts vertically so that speed will stay constant). In the horizontal equation, there is no acceleration.

Now, you are given the acceleration due to gravity on the moon and you know there was no initial vertical velocity (you are aiming horizontally) so you can calculate how far downward the dart will go. For the vertical equation, the initial velocity is 0.

If that is less than 1.85 meter height, you are finished- that is the distance below the teacher's nose the dart hit.

If that is more than 1.85 m, your dart did not reach the target. Go back and calculate the time it took for the dart to drop (vertically) 1.85 meters (when it hits the ground). Use that time to find how far horizontally the dart will go. Since the problem asks "how far in front of the target", you will need to subtract that from the 15 meter distance.

 

[learningphysics]

My approach is to get the time it takes to hit the target (travel 15 m). Also get the time it takes to hit the ground.

See which is less. If time to hit the target is less, then use that time to get the distance below line of sight.

If time it takes to hit the ground is less, then use that time to calculate how far in front of the teacher the dart hit.

 

[mohlam12]

here is what i did (on other moon, g= 0.074 m/s/s)
the time the dart will go horizontally is 0.578 sec before it hit the target and vertically the dart will go 0.523sec before it hit the ground. so, the the dart will not hit the teacher, however it will hit the ground 1.428m in front of the teacher.
0.578-0.523=0.055 sec (if it didn t hit the ground, the dart will tale 0.055 sec to hit the teacher)
then d= 0.055 * 25.97 (it is the velocity of the dart horizontally)
=1.428 m

but if i used hallsofivy procedure, i got 0.012 m in front of the teacher ??

 

[learningphysics]

here is what i did (on other moon, g= 0.074 m/s/s)
the time the dart will go horizontally is 0.578 sec before it hit the target and vertically the dart will go 0.523sec before it hit the ground. so, the the dart will not hit the teacher, however it will hit the ground 1.428m in front of the teacher.
0.578-0.523=0.055 sec (if it didn t hit the ground, the dart will tale 0.055 sec to hit the teacher)
then d= 0.055 * 25.97 (it is the velocity of the dart horizontally)
=1.428 m

but if i used hallsofivy procedure, i got 0.012 m in front of the teacher ??

Hmmm... that's strange. Halls told you to get the time it takes to hit the ground, and calculate the distance in front of the teacher... same thing I said. So you should have followed the same procedure both times...

Are you doing this for two different moons?
At one point you say
g= 0.074 m/s/s
and previously
g=1.624

Are you still using a height of 1.85m?

d=0.5gt^2 (vertically)
If I solve:
0.5(1.624)t^2=1.85
I get t=1.509s > 0.578s so it hits the target.

If g=0.074
0.5(0.074)t^2=1.85
I get t=7.07s > 0.578s so again it hits the target.

Maybe I'm misunderstanding something.

 

[mohlam12]

yup, well i took the moon just for an example because it s the most common moon...this scene happened in Himalia, it s a Jupiter moon, which i found its gravity to be 0.074 m/s/s

 

[learningphysics]

Can you show how you got 0.523s?

 

[mohlam12]

ok yeah now, the dart will go (7.07-0.578= 6.49 sec ; 6.49 * 25.97 = 168.54 m) 168.54m past the teacher. now, the problem says :
your requirement is to determine within 5 cm how far below the horizontal line of sight will you hit?
i don't know anything about the line of sight, or maybe I didn't get what I have to do now! yeah, i need help :-D

 

[learningphysics]

ok yeah now, the dart will go (7.07-0.578= 6.49 sec ; 6.49 * 25.97 = 168.54 m) 168.54m past the teacher. now, the problem says :
your requirement is to determine within 5 cm how far below the horizontal line of sight will you hit?
i don't know anything about the line of sight, or maybe I didn't get what I have to do now! yeah, i need help :-D

So the dart travels for 0.578 s. How far does it travel vertically in that time? That's the distance below the line of sight. Don't worry about the words "line of sight"... All that means is at the height of the teacher. Remember your muzzle is also at this same height.

So it is asking how far below that height does it hit the teacher. in other words how far downward does it travel?

 

[mohlam12]

well, the 0.523 sec is wrong, 0.523 should be the final velocity sqrt(2*0.074*1.5) yhrn if u divide it by 0.074, u get 7.07 sec (the right answer) BUT now i dun know what i have to do according to the problem, i didn t get the question ?!

 

[mohlam12]

oooh okay, now i think i got it

 

[learningphysics]

well, the 0.523 sec is wrong, 0.523 should be the final velocity sqrt(2*0.074*1.5) yhrn if u divide it by 0.074, u get 7.07 sec (the right answer) BUT now i dun know what i have to do according to the problem, i didn t get the question ?!

Ok... you've got the 7.07 s which is greater than 0.578 s.

So you know it hits the target.

Now you don't need the 7.07 s anymore... you just needed that to see if it hits the target or not. Now you know it hits the target.

You know it takes 0.578 seconds to hit the target.

In 0.578 seconds, what is the distance traveled vertically?

 

[mohlam12]

wait, but why is there the 5cm in the problem. , i didn t get it!
i got that the dart will travel 0.012 sec in 0.578 sec! what is the 5 cm for?

 

[learningphysics]

HallsOfIvy's approach is a little quicker.

 

[learningphysics]

wait, but why is there the 5cm in the problem. , i didn t get it!
i got that the dart will travel 0.012 sec in 0.578 sec! what is the 5 cm for?

That's for rounding.

You got 0.012m?

Convert that to cm. And then round to the nearest 5cm (in other words 0cm or 5cm or 10 cm etc...).

 

[mohlam12]

oh okay, i see. thank you for your help!

 

[learningphysics]

oh okay, i see. thank you for your help!

OOPS! No don't round it!

I misread. They want it accurate to within 5cm.

So your answer of 0.012m = 1.2 cm is fine

I thought they wanted it to the nearest 5cm but that's not what it said.

 

[mohlam12]

okay! thx,