Motion in two dimensions question. At wit's end.

Click For Summary

Homework Help Overview

The discussion revolves around a projectile motion problem where the original poster is trying to determine the minimum time of flight for a projectile launched at an initial speed of 80 m/s to hit a target 450 m away. The problem is complicated by the absence of a specified launch angle.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses frustration over the lack of an angle, leading to difficulties in eliminating variables. They suggest that the solution involves finding the smallest angle for the projectile to reach the target distance.
  • Some participants suggest considering the time of flight in relation to the projectile's upward and downward motion, indicating that the total time must account for both phases.
  • Others propose writing equations for both horizontal and vertical positions to eliminate the angle and solve for time.

Discussion Status

Participants are actively engaging with the problem, offering various approaches to help the original poster. There is a sense of progress as the original poster indicates they will attempt the problem again based on the suggestions received.

Contextual Notes

The original poster mentions having the answer from the textbook but struggles to derive it independently, indicating a potential gap in understanding the application of projectile motion equations without a specified angle.

Hekmat123
Messages
3
Reaction score
0
[SOLVED] Motion in two dimensions question. At wit's end.

Hi, I'll try to keep this short. I have a Physics exam tomorrow and was doing some textbook questions and can't figure out this one problem. It's killing me! I have the answer (supplied by the back of the book), but just can't seem to get there.

Some help would be much appreciated since I've grown very frustrated with it.


Homework Statement



The initial speed of a projectile is 80 m/s. If the projectile is to strike a target that is a horizontal distance of 450 m away, what is the minimum time of flight. (Answer 6.1s)

Homework Equations



Standard laws of motion ignoring air resistance and the like

X(final) = X(initial) + V(initial) (t) + 0.5at^2


The Attempt at a Solution



I've tried everything I can think of. What's shutting me down every time is that there is no angle supplied so I'm always stuck with two variables I can't eliminate.

Obviously the answer must lie in finding the smallest angle possible for a projectile at 80 m/s to reach 450 meters, I just can't figure out how. I came up with this equation from the formula above, but I have no clue how to use it as my brain is just shutting down at this point (maybe it's useless i don't know, argh!)

450 = 0 +80(cos(alpha))t = 0
t= 5.625/cos(alpha)



If someone could help me get this problem you would earn my undying admiration, and I'll mail you a cookie!
 
Last edited:
Physics news on Phys.org
Take a step back. The target is 450m away, so the time it takes to get there is what you calculated in terms of alpha. The other constraint is that the projectile has to go up for half of the flight, and back down for the other half of the flight, in order to hit the target. That up and down motion has to take the same total time t.

You'll probably get a 2nd order equation, since there are probably two answers (low angle shot < 45 degrees, and high angle shot > 45 degrees)
 
You already have an equation for the horizontal position as a function of time. Now write an equation for the vertical position as a function of time. Combine those equations to eliminate the angle and solve for the time.
 
Thanks for the help! I'm going to try it again and I'll post back, hopefully to add a solved tag , heh.
 
Finally managed it. Thanks for the help guys. All I had to do was use the sin/cos identities to solve for alpha, I feel retarded, thanks again!
 

Similar threads

  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 20 ·
Replies
20
Views
11K
Replies
4
Views
2K
Replies
12
Views
5K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K