2 Projectiles are fired simultaneously

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Two projectiles are launched simultaneously from ground level with the same initial speed and cover a horizontal distance of 160m, but one lands 6 seconds before the other. The motion is governed solely by gravitational acceleration, g=10m/s². The angles of projection for the two projectiles are complementary, with one at theta and the other at 90 - theta. The participant attempted to equate their ranges and times but struggled with the resulting trigonometric equations. Further assistance is requested to solve the equations derived from the problem.
Om Swostik

Homework Statement


2 projectiles are fired simultaneously from ground level with same initial speed (u).Both cover same horizontal distance of 160m on reaching the ground level .One of them reaches 6 sec prior to the other.Only gravitational acceleration g=10m/s squared governs the motion of both the projectile.Calculate u.[/B]

Homework Equations


Range(horizontal distance)=u squared sin 2×theta/g= Ux cos theta ×time
Time=2u sin theta/g
Where ,
U is initial velocity
Theta is the angle of projection
and T is the time taken by the projectile to reach ground level.
Note:vector u has 2 components which are Ux=u cos theta and Uy=u sin theta

The Attempt at a Solution

 
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Om Swostik said:

Homework Statement


2 projectiles are fired simultaneously from ground level with same initial speed (u).Both cover same horizontal distance of 160m on reaching the ground level .One of them reaches 6 sec prior to the other.Only gravitational acceleration g=10m/s squared governs the motion of both the projectile.Calculate u.[/B]

Homework Equations


Range(horizontal distance)=u squared sin 2×theta/g= Ux cos theta ×time
Time=2u sin theta/g
Where ,
U is initial velocity
Theta is the angle of projection
and T is the time taken by the projectile to reach ground level.
Note:vector u has 2 components which are Ux=u cos theta and Uy=u sin theta

The Attempt at a Solution

An attempt is required. Forum rules.
 
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THE ATTEMPT AT A SOLUTION:
I took the angle of projection of 1st projectile as theta and the angle of the 2nd one as 90 - theta (as there can be two angles of projection for the same range).
I equated the range and the time ,that is
T=t+6.As I have really limited knowledge of trigonometry, the resulting equation which I arrived at couldn't be solved by me.
 
Om Swostik said:
THE ATTEMPT AT A SOLUTION:
I took the angle of projection of 1st projectile as theta and the angle of the 2nd one as 90 - theta (as there can be two angles of projection for the same range).
I equated the range and the time ,that is
T=t+6.As I have really limited knowledge of trigonometry, the resulting equation which I arrived at couldn't be solved by me.
Please post the equation. It might save time to post your working too, in case you have a mistake somewhere.
 
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?

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