One problem, three checked solution families
A launcher is 1.25 m above level ground. Fired horizontally, a ball lands 5.0 m from the point directly below the launcher. Ignore air resistance and use g = 10 m/s^2. The same launcher speed is then aimed vertically upward. How high above the launcher does the ball rise?
Reveal the checked result
5 m. In the interactive lab, the result stays hidden until the numeric gate passes.
What counts as a different solution?
A new family must change the load-bearing mathematical dependency path. Rewording, rearranging equations, or renaming the same method does not increase the count.
Open a solution passport only when you are ready for the route.
Time → speed → vertical rise
From 1.25 = (1/2)(10)t^2, obtain t = 0.5 s. From R = ut, obtain u = 5.0/0.5 = 10 m/s. At the top of the vertical launch, 0 = u^2 - 2gH, so H = 5.0 m.
Trajectory / direct elimination
For the horizontal launch, h = gR^2/(2u^2). For the vertical launch, H = u^2/(2g). Eliminating u and g gives hH = R^2/4, hence H = R^2/(4h) = 5.0 m.
Recover speed, then use energy
The horizontal fall gives t = 0.5 s and u = 10 m/s. For the vertical launch, (1/2)mu^2 = mgH. Cancel m: H = u^2/(2g) = 5.0 m.