question projectile motion (1 Viewer)

[richz]

i got a question
An inexperienced shooter aims a rife horizontally at a target 300.0m away. if the bullet leaves the barrel of the gun at 2500m/s, how far below the centre of the target will the bullet strike??

anyone have any ideas, even though i THINK questons like this wont be asked in the new syllabus. if it is just post reply and if you can help me out also post a reply

 

[acmilan]

I dont see why it cant be asked in the HSC.

s_x = 300 m
u_x = 2500 m/s
u_y = 0 m/s
s_x = ut
300 = 2500t
t = 0.12 secs

s_y = u_yt + 1/2at²
s_y = 0 + 1/2*9.8*(0.12)²
s_y = 0.07056 m

So it will land 0.07056 m or 7.056 cm below target

 

[Xayma]

i got a question
An inexperienced shooter aims a rife horizontally at a target 300.0m away. if the bullet leaves the barrel of the gun at 2500m/s, how far below the centre of the target will the bullet strike??

anyone have any ideas, even though i THINK questons like this wont be asked in the new syllabus. if it is just post reply and if you can help me out also post a reply

g=-9.8ms^-2

Now the horizontal component will always be the same. Hence it will take 300/2500 seconds to hit 0.12 secs.

Now in that time it will move down:
s=0-1/2*9.8*.12²
ie it will land 7.056cm below the centre.

 

[Numero Uno]

lol its good too see u both agree on the answer

and i will also agree to both or ur answers

 

[acmilan]

Haha beat him by 1 minute

 

[richz]

thnak you very much guys!!

 

[richz]

Guys, i have trouble working out projectile motion questions, does anyone have tips on working them out??

 

[Numero Uno]

read the qustion carefully everytime
find out what exactly they want u to find out

and PRACTICE PRACTICE !!

 

[acmilan]

Normally if im not in a rush i write down all the mathematical information given then see which formula can be used in which there is enough information. After you do lots of questions you will see that most have similar trends and will be easy to see what approach to take.

 

[richz]

ok heres another one i dont understand

A stone is thrown a distance of 75m at an elevation of 24.3 degrees, calculate its original speed?

 

[aim54x]

is there more data??

 

[richz]

nope not by the question on my sheet

 

[physician]

nope not by the question on my sheet

Its either a brain draining question.. or the other possibilty is that u've left out a vital part of the question...

just repost ur question....

or it could be that the holiday has drained out some of that physics problem solving poison.....

 

[Xayma]

It is possible, just solve the set of simulatenous equations you will generate.

You know that:

75=v*cos 24.3^o*t
vt=75/cos 24.3^o

solve for r_y=0 to get a formula for t, solve.

 

[acmilan]

It is possible, just solve the set of simulatenous equations you will generate.

You know that:

75=v*cos 24.3^o*t
vt=75/cos 24.3^o

solve for r_y=0 to get a formula for t, solve.

Tried it that way and got this. Dont know if method is right and if it is theres a chance of approximation error because i used the calculator on my computer.

edit: this method assumes the stone landed on the same plane that it was thrown

 

[wanton-wonton]

read the qustion carefully everytime
find out what exactly they want u to find out

and PRACTICE PRACTICE !!

No shit Einstein.

 

[wanton-wonton]

g=-9.8ms^-2

Now the horizontal component will always be the same. Hence it will take 300/2500 seconds to hit 0.12 secs.

Now in that time it will move down:
s=0-1/2*9.8*.12²
ie it will land 7.056cm below the centre.

No, g is not -9.8. g is only -9.8 when you're throwing something against gravity. In this case, the bullet is falling, so g is 9.8.

 

[Xayma]

I assumed that acceleration upwards was positive as is standard. Hence g is a negative as it is acceleration downwards.

 

[wanton-wonton]

Yeah...but...you're throwing the object against gravity, so hence negative. In this case, it's falling towards gravity, same direction, hence positive.

 

[acmilan]

Doesnt make a difference as long as you are consistent