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## Main Question or Discussion Point

I've been trying to determine the optimun angle of launch 1,5 metres above the ground by pooceeding from the basic equations for throwing motions:

x=vcosθ*t

y=y0+vsinθ*t-gt^2/2, y0=1,5m, g=9,8m/s^2

What I've done so far is expressing t as x/vcosθ. Insertion in the equation for y gives:

y=y0+xvsinθ/vcosθ-(g/2)(x/vcosθ)^2

y equals 0 when the object hits the ground. This means:

0=y0+xvsinθ/vcosθ-(g/2)(x/vcosθ)^2

I've then expressed x as a function of θ by solving the quadratic equation above.

It's desirable to derive the function and then solve the equation x'(θ)=0 to find what angle that gives the maximum range 1,5 metres above the ground.

I have however been unsuccesful so far and would be very grateful for any advice regarding how to derive the function and solving the following equation. I'm not that advanced in physics and would therefore apreciate thorough explanations.

Please help me out here, because I'm totally lost.

x=vcosθ*t

y=y0+vsinθ*t-gt^2/2, y0=1,5m, g=9,8m/s^2

What I've done so far is expressing t as x/vcosθ. Insertion in the equation for y gives:

y=y0+xvsinθ/vcosθ-(g/2)(x/vcosθ)^2

y equals 0 when the object hits the ground. This means:

0=y0+xvsinθ/vcosθ-(g/2)(x/vcosθ)^2

I've then expressed x as a function of θ by solving the quadratic equation above.

It's desirable to derive the function and then solve the equation x'(θ)=0 to find what angle that gives the maximum range 1,5 metres above the ground.

I have however been unsuccesful so far and would be very grateful for any advice regarding how to derive the function and solving the following equation. I'm not that advanced in physics and would therefore apreciate thorough explanations.

Please help me out here, because I'm totally lost.