How come we don’t have to consider the extra distance travelled since the ball was lauched at a slight angle? (10+s) where s is the distance between the max altitude and horizontal?
The equation s = 5t – 10t²/2 takes into account the fact that the ball is originally going up at 5m/s.
If you plot s against time you find that the ball reaches a maximum height at t = 0.5, and at that time s is 1.25m. After that it starts going downwards and after 2 seconds it ends up at -10m.
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