Answer:
26.14 ft/s
Step-by-step explanation:
Assume that the acceleration was constant throughout the whole fall and equal to the acceleration of gravity which is roughly 32.17 ft/s.
If the distance travelled was 26 ft, the time it took him to land was:
[tex]s(t) = 16t^2\\26 = 16t^2\\t=1.27475/ s[/tex]
At the apex, just before the fall, the skateboarder velocity is zero. The velocity of an object, initially at rest and accelerating at a constant rate, is:
[tex]V =a*\frac{t^2}{2}\\V= 32.17\frac{1.625}{2}\\V=26.14\ ft/s[/tex]
The skateboarder's velocity at the moment he touched down onto the ramp was 26.14 ft/s.