Projectile Motion -- Help please understanding these basic problems

In summary, the range on a horizontal plane is 122.5 meters when the angle of projection is 15°. The error in elevation at this range is ± .7.6 meters.
  • #1
RandomStudentNotused
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Homework Statement
Hello. Recently in my class we started Projectile motion but I seem to have some trouble. Can you help me with these questions?
Relevant Equations
v=u+at
vSinθ
vCosθ
S=ut
v=ut+1/2at²
V²=U² + 2as
1)Find the range of the gun when the muzzle speed is 400m/s and the elevation 24.5°

- I tried to find " Time " via t= [v-u]/a by substituting it through cos-->

t= ( 0- [400cos24.5] )/-10 and got [ 36.39 ].

Which I then substituted in S=ut---> S= ( 400cos24.5 ) x 36.39=13245.3.

[ the answer in the answer key is 10.9 km ]

2)A target is 10 km from a gun and the gun's muzzle speed is 400 m/s. What angle of projection. What angle of projection will ensure a hit?

-- I used cos-1 (10000/400 ) and got cos-1(25). I think its wrong.

So I tried θ = 400sin24.5=165.877

[ the answer in the answer key is 71° 5' ; 18° 55' ]

3)A projectile has a muzzle speed of 70 m/s. what is the range on a horizontal plane if the angle of projection is 15° . Find the error, at this range if the error in the elevation is 0.25°

- I used t = v - u/ a and substituted the values in T= ( 0 - 70sin15 ) / -10=1.81

And using formula S=ut, I multiplied it--->S= ( 70cos15 ) x 1.81=122.5 m[ the answer in the answer key is 250m ; ± .7.6 m ]

-----

PS: Can you tell me, when do you use cos and sin in questions like these?

-do you use sin when something goes up or is in the air? And cos for horizontal range of a projectile/object?
If you can help,it would be much appreciated
 
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  • #2
Hello student, @RandomStudentNotused (?)
:welcome: !​

RandomStudentNotused said:
1)Find the range of the gun when the muzzle speed is 400m/s and the elevation 24.5°

- I tried to find " Time " via t= [v-u]/a by substituting it through cos-->

t= ( 0- [400cos24.5] )/-10 and got [ 36.39 ].

So it looks as if you think the horizontal velocity (*) at the end (your symbol ##\ v\ ##) will be zero. If you make a sketch, that can be seen to be very unlikely :rolleyes:

(*) or did you mean vertical :wink: ? Check ##\sin## and ##\cos## definitions !

Free tips:
  • Make a sketch
  • Fill in the known / given data as symbols
  • Mark the variable / quantity you are looking for
  • Search for appropriate variables on the path from given to searched for, and write down the relationships in terms of symbols
  • Check dimensions
  • Do the math
  • Check the answer
  • Check the answer
You're doing several of them already, so you're on the right path!
 
  • #3
RandomStudentNotused said:
PS: Can you tell me, when do you use cos and sin in questions like these?
My only advice on this is that you use the cosine when it's the cosine, and you use the sine when it's the sine! And if you get them mixed up now and again, then you're in good company.
 
  • #4
Your working suggests you don't really understand what you are doing (sorry)!

At the risk of being accused of self-promotion, try this video:


(Edit -typo's corrected.)
 
  • Like
Likes Lnewqban and BvU
  • #5
Steve4Physics said:
At the risk of being accused of self promotion
Haha, on the contrary ! A real treasure trove of didactics ! Well done !
 
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