Projectile motion problem - velocity , angle, and vertical acceleration unknown

AI Thread Summary
The discussion revolves around solving a projectile motion problem on a foreign planet where gravity is not equal to 9.8 m/s². Key parameters provided include a maximum horizontal distance of 40.2 m reached in 25 seconds and a maximum vertical distance of 25 m in 12.5 seconds. The equations of motion are established to find the initial velocity, launch angle, and vertical acceleration, with three main equations derived from the vertical and horizontal components of motion. The participants focus on eliminating variables to solve for the unknowns, emphasizing the relationships between vertical and horizontal motion. The conversation highlights the complexity of the problem while seeking collaborative input for a solution.
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This is all the given info:
On a foreign planet (g cannot equal 9.8m/s), a profectile is fired

Maximum horizontal distance= 40.2 m and it reaches this distance at 25s
Maximum vertical distance= 25 m and it reaches this distance at 12.5s

There is no horizontal acceleration, and there is no mass given (and I'm pretty sure its not needed to find the

solution)

What is the initial velocity of the projectile and angle at which it was fired and what is the acceleration due to

gravity

So basically I need to find vertical acceleration, intitial velocity, and angle theta.




I'm really stumped on this and all my attempts at a solution was just me listing out the givens.
 
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What equations do you have?
 
So let's see... If this thing lands at the same vertical height it was launched at then (Vsin theta)^2/(2*y_max)="g" and the horizontal distance is Vi^2*sin(2*theta)/"g" and V*cos(theta)=distance in x/time. 3 equations, 3 unknowns.

Anyone want to look at this problem?:
https://www.physicsforums.com/showthread.php?t=357356
 
Let the initial speed be v, the angle \theta and acceleration due to gravity a.

For the vertical rise, use x = (v\sin{\theta})t+\frac{1}{2}at^2, where v\sin{\theta} is the vertical velocity.

25 = v\sin{\theta}(12.5)+\frac{1}{2}a(12.5)^2

\Rightarrow 25 = 12.5 v\sin{\theta}+78.125a --------(1)

Since the time it takes to fall back down is 25 seconds, we also have

0 = v\sin{\theta}(25)+\frac{1}{2}a(25)^2

\Rightarrow 0 = 25v\sin{\theta}+312.5a --------(2)

For the horizontal distance, use x = (v\cos{\theta})t (a = 0 in the horizontal component)

40.2 = v\cos{\theta}(25) --------(3)


From equations (1) and (2), we can eliminate a and v\sin{\theta} (or just solve for a)

v\sin{\theta} = 4 --------------(4)

From equation (3),

v\cos{\theta} = 1.608 ----------(3')

To find \theta, divide (4) by (3')

To find v, square (4) and (3'), add them together, and use a trigonometric identity to simplify
 
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