Projectile launch point problem

Click For Summary
A projectile is launched at 26.0 m/s and 40.0 degrees, hitting the ground after 8.00 seconds. The launch point is determined to be 180 m lower than the landing point. The user struggles with calculating the maximum height above the launch point, facing issues with multiple unknowns in their equations. A suggestion is made to use the formula v_{f}^{2}=v_{i}^{2}+2ax to find the maximum height. The user successfully resolves their issue with the provided hints.
SA32
Messages
31
Reaction score
0
I'm having some difficulty with a projectiles question.

"A projectile is fired with an initial speed of 26.0 m/s at an angle of 40.0 degrees above the horizontal. The object hits the ground 8.00 s later."

The first part of the question is: "How much higher or lower is the launch point relative to the point where the projectile hits the ground?
Express a launch point that is lower than the point where the projectile hits the ground as a negative number."

Which I already found out using y=(viy)(t)-(1/2)(g)(t^2)

Substituting the given numbers, I get -180 m, which would be the case if the landing point was below the launch point. Since the question wants a launch point lower than the landing point, the answer is 180 m.

The second part, where I'm having trouble: "To what maximum height above the launch point does the projectile rise?"

What I have,

x=(26)(cos(40))(t)
y=(26)(sin(40))(t)-(1/2)(g)(t^2)

I can't substitute 180 m or 8 s because those numbers apply to the landing point, and I'm looking for ymax so I end up with three unknowns in two equations, which I cannot solve. I don't know how to set this up so that I can solve it.

Thanks for any help!
 
Last edited:
Physics news on Phys.org
To find maximum height, remember that the vertical component of velocity at the maximum height is zero.

Does this help?
 
Hint: Try using the formulae;

v_{f}^{2}=v_{i}^{2}+2ax

:wink:

Edit: Beaten to it.
 
Whoops! Should post here to let you know I used your hints to figure it out, so thanks!
 
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
40
Views
2K
  • · Replies 15 ·
Replies
15
Views
26K
Replies
23
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 39 ·
2
Replies
39
Views
4K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
8
Views
2K
Replies
23
Views
3K
Replies
25
Views
3K