Projectile Problem Homework: Solve Range, Time, Velocity, Height

  • Thread starter Thread starter munkachunka
  • Start date Start date
  • Tags Tags
    Projectile
Click For Summary

Homework Help Overview

The discussion revolves around a projectile motion problem involving a mortar shell fired at an angle of elevation. The original poster seeks to determine various parameters such as time of flight, initial velocity, range, and height at specific intervals, while neglecting air resistance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to rearrange equations related to projectile motion to find time of flight and initial velocity. They express uncertainty about the calculations for the height and range at a specific time.

Discussion Status

Participants have provided feedback on the original poster's calculations, noting that the initial parts appear correct but suggesting a re-evaluation of part 4. Some participants have pointed out potential typos and slight discrepancies in numerical results, indicating a collaborative effort to clarify and verify the work presented.

Contextual Notes

There is mention of specific marks allocated to part 4, which raises questions about the complexity of that part compared to the others. The original poster expresses concern about the perceived simplicity of part 4 relative to its weight in the overall problem.

munkachunka
Messages
36
Reaction score
0

Homework Statement



I have made an attempt at the following could someone please check my work.

a mortar shell fired with an angle of elevation of 52 deg has a range of 2.1km, neglecting air resistance, determine.

1, time of flight
2, initial velocity
3, shells range and height after 10 secs
4, the other range at which the height was the same as calculated in part 3 above

Homework Equations



x=u(cos(angle))t
h= u(sin, (angle))t-1/2gt^2

The Attempt at a Solution



a, rearanging x=u(cos52)t to equal x/cos52=ut
1hen x = 2.5km h=0

h=u(sin angle)t-1/2gt^2

substituting 0=(2.5knm/cos52)*(sin52)-1/2.9.8.t^2

transposing for t = 23.409

2, x=u(cos52)t therefore u=2100/(cos52)*23.409 = 145.714m/s

3= x=u(cos52)t = 145.712(cos52).10 = 897.093M

height = u(sin52)t-1/2gt^2
= 145.712*(sin52)*10-1/2*9.8*10^2

height = 658.226

4 assuming angle of firing = angle of descent and initial acceleration = rate of deceleration then the other range = 2.1km-897.093 = 1202.907

assuming the first 3 parts are correct then the part I am not too sure about is part 4, there are 12 marks allocated to this and so looks too easy.

your time is greatly appreciated
 
Physics news on Phys.org
munkachunka said:
assuming the first 3 parts are correct then the part I am not too sure about is part 4, there are 12 marks allocated to this and so looks too easy.
It looks good to me. (Just round off your answers to a reasonable number of significant figures.)

If you are unsure about part 4, solve it a second way and compare. (Use your equation for height as a function of time, solve for the time, plug into your equation for x to find the range.)
 
Yeah, everything looks good... Just noticed this part:

0=(2.5knm/cos52)*(sin52)-1/2.9.8.t^2

the problem says 2.1km, you left a t out (probably nothing to do with your written work, but just typos)... but your calculations are right...

I noticed I was getting a few decimal places off from you in the answers... but other than that everything was the same.
 
Last edited:
thankyou for your time
 

Similar threads

Replies
40
Views
3K
Replies
4
Views
2K
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
6
Views
3K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K