A skateboarder free-fell 26 ft onto a ramp. The distance s(t), in feet, traveled by a body falling freely from rest in t seconds is approximated by s (t )equals 16 t squared. Estimate the skateboarder's velocity at the moment he touched down onto the ramp.

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:

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:

The skateboarder's velocity at the moment he touched down onto the ramp was 26.14 ft/s.

idk

step-by-step explanation:

idk

y=3x+4

step-by-step explanation:

given the points a(0,4) and b(-4, -8)

eqn of line ab: (y-4)/(x-0)=(-8-4)/(-4-0)

(y-4)/x=-12/-4=3

y=3x+4