Question

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.

Answer 1

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:

$s(t) = 16t^2\\26 = 16t^2\\t=1.27475/ s$

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:

$V =a*\frac{t^2}{2}\\V= 32.17\frac{1.625}{2}\\V=26.14\ ft/s$

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