Object lands higher than launch: trajectory ends at known value of y, find flight time and range.

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Flight time and range for a projectile flight in which the object lands higher than launch height.
We start by finding the launch velocity x and y components. The horizontal velocity component is given by speed*cos(theta) and the vertical velocity component is given by speed*sin(theta).
Next, we attack the flight time, uneven surface problem. Because the trajectory ends at known value of y, we use the y equation of motion to start the analysis. Plugging in the known final y value, we obtain a quadratic equation for the flight time. Solving the equation, we obtain two physical values of t: one for the way up, and one for the landing spot on the elevated surface (this is the time we keep).
Next, we plug the flight time into the x equation to get the range of the projectile. Since the x acceleration is zero in projectile motion, this equation is simple: x=x_0+v_0t. Plugging in the initial position of zero and the known initial x velocity component, we find the range of the projectile.

[@Mike-ks6qu]

Amazing. I was stuck on a very similar problem. Thanks so much. 👏

[@ZaksLab]

glad I could help! z

[@gyarafin1204]

thank you zak

[@ZaksLab]

you're welcome! z