Finding Initial Angle for a Projectile to Hit a Target 10,000 ft Away

Click For Summary
To find the elevation angles for a projectile to hit a target 10,000 ft away with a speed of 800 ft/s, two equations are used: one for vertical motion and one for horizontal motion. A key mistake identified was in solving for sin(θ) in the vertical motion equation, which should be sin(θ) = t/50 instead of the previously stated form. The correct approach involves eliminating time (t) from the equations to derive the angles. One angle has been confirmed to match the answer key, while the second angle remains elusive. Accurate representation of work is encouraged for clarity in problem-solving.
Marcin H
Messages
306
Reaction score
6

Homework Statement

:
[/B]
Find two elevation angles that will enable a shell, fired from

ground level with a muzzle speed of 800 ft/s, to hit a groundlevel

target 10,000 ft away.

Homework Equations


Look at attached picture

The Attempt at a Solution


Look at attached picture

***I get to cos(x)sin(x)=1/4, but I don't think you can solve that for an angle x...
 

Attachments

  • New Doc 1_1.jpg
    New Doc 1_1.jpg
    40.8 KB · Views: 578
Physics news on Phys.org
***UPDATE***

I got one of the angles, and it checks out with the answer key, but how on Earth do I get the angle of 15?

MY WORK:
New Doc 2Page 1.jpg

ANSWER KEY:
ANSWER KEY.png
 
Marcin H said:
***UPDATE***

I got one of the angles, and it checks out with the answer key, but how on Earth do I get the angle of 15?

MY WORK:
View attachment 89769
ANSWER KEY:
View attachment 89770
You have a mistake in your work. As best as I can tell, your answer is close only due to a coincidence.

One of your equations is ##50\sin(\theta) = t## and the other is ##800 \cos(\theta)t = 10000##
Your mistake is in solving for ##\sin(\theta)## in the first equation. It should be ##\sin(\theta) = \frac t {50}##.

BTW, please make an effort to post your work here, rather than as an image. Your work is somewhat difficult to read.
 
S=ut-½at2. (1)
R=vt. (2)

Just eliminate t, and substitute data in the last equation.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
10
Views
4K
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K