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I have been trying to figure out how to solve it.

I know how to get the velocity by the distance and how to get the max height.

Distance: $d = V_0^2 \cdot \sin(2\cdot\alpha)/g$

Max height: $h_{max} = V_0^2 \sin(\alpha)^2/(2\cdot g)$

so using the $h_{max}$ to substitute $V_0^2$ I think would be: $$V_0^2 = \frac{h_{max}}{\frac{\sin(\alpha)^2}{2\cdot g}} $$

so if I substitute this on the distance equation, I think it would be: $$ d = \frac{h_{max}}{\frac{\sin(\alpha)^2}{2\cdot g}} \cdot \frac {\sin(2 \cdot\alpha)}{g} $$

But I don't know how to solve the $\sin$ to get the angle and even I don't know if this equation is ok.

For example: I want a max height = 8 meters and a distance 5 meters, I would like to know what angle I have to use, with that angle I can get the force I need.

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The key concept for projectile motion problems is this: what happens in the vertical direction is mathematically COMPLETELY independent of what happens in the horizontal direction. This means that you can actually work this type of problem as two separate problems by following these steps:

  1. You want a maximum height of 8 meters. Pretend that you are launching your object vertically upwards to this height, and use kinematic equations in the vertical direction only to calculate the launch velocity and how long the object will be in the air (the object has to rise to 8 meters, then fall back to its launch point).

  2. Using the time in step "1", and the desired horizontal displacement of 5 meters, pretend that you are launching your object horizontally, that the object will continue moving in the horizontal direction only, and that the object will maintain a constant horizontal speed (i.e., gravity does NOT act horizontally, so it can't change the horizontal speed of the object). What horizontal velocity must the object be launched at to travel 5 meters in the calculated time?

  3. You now have the vertical velocity component and the horizontal velocity component of your object. Add these components via vector addition to calculate the actual launch velocity and the required angle.

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  • $\begingroup$ Thank you so much, it is working and it was more easier than I thought. $\endgroup$ – alexbg Mar 14 at 10:16

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