What Is the Optimal Launch Angle for Maximum Range When Landing 3 Meters Lower?

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To determine the optimal launch angle for maximum range when landing 3 meters lower, it is essential to derive equations for both horizontal and vertical motion. The standard launch angle for maximum range is 45 degrees when the launch and landing heights are equal, but adjustments are needed for different heights. The discussion emphasizes using differentiation to maximize the range function based on the given initial velocity of 5 m/s. Participants suggest writing simultaneous equations for horizontal and vertical motion to solve for the final coordinates. Clarification is sought on the specific launch and landing heights to ensure accurate calculations.
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How can i find the angle of which an object will be thrown the farthest?

I know the angle will be 45 degrees if it were to be catched at the same height it were thrown, but what I'm looking for is the angle if it were 3meters off the ground.

Thank you for the help.
 
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larryboi7 said:
How can i find the angle of which an object will be thrown the farthest?

I know the angle will be 45 degrees if it were to be catched at the same height it were thrown, but what I'm looking for is the angle if it were 2meters off the ground.

Thank you for the help.

Welcome to the PF. Are you familiar with how to maximize a function using differentiation? Write an equation for the range as a function of launch angle (given your offset y information). Show us your attempt at a solution so we can help.
 
berkeman said:
Welcome to the PF. Are you familiar with how to maximize a function using differentiation? Write an equation for the range as a function of launch angle (given your offset y information). Show us your attempt at a solution so we can help.

The equation would be

initial velocity is 5m/s

0 = 3 + 5sin(x) - (1/2)gt^2

x(t) = 5cos(x) * t

I know what differentiation is but I can't see how it will work to find the angle for maximum range.
 
larryboi7 said:
The equation would be

initial velocity is 5m/s

0 = 3 + 5sin(x) - (1/2)gt^2

x(t) = 5cos(x) * t

I know what differentiation is but I can't see how it will work to find the angle for maximum range.

I'm not understanding what you wrote, but it sounds like you can figure this out. BTW, is it thrown at 0m and caught at 2m, or the other way around?

I think you will write two equations (this is the usual way to solve this type of question): One for the horizontal motion, which has a constant velocitty. The other is the vertical motion equation, which involves the vertical pull of gravity. You will use them as simultaneous equations, and solve them for the final simultaneous (x,y) point. Does that help? Show us your work...
 
berkeman said:
I'm not understanding what you wrote, but it sounds like you can figure this out. BTW, is it thrown at 0m and caught at 2m, or the other way around?

I think you will write two equations (this is the usual way to solve this type of question): One for the horizontal motion, which has a constant velocitty. The other is the vertical motion equation, which involves the vertical pull of gravity. You will use them as simultaneous equations, and solve them for the final simultaneous (x,y) point. Does that help? Show us your work...

it is thrown at 3m and and caught at 0 meters.
i have the equations as stated before. would you mind showing me?
 
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