When an object is dropped, the distance it falls in t seconds, assuming that air resistance is negligible, is given by s(t)equals=4.9054.905t2 where s(t) is in meters (m). If a stone is dropped from a cliff, assuming that air resistance is negligible, find the velocity and acceleration of the stone after it had been falling for 1515 sec.

velocity = 147.15 m/s

Acceleration = 9.81 m/s²

Step-by-step explanation:

s(t )= 4.905t²

the  velocity, ds / dt = 2 * 4.905 t

= 9.81 t

if

t = 15 secs

velocity, which is

ds / dt = 9.81 * 15

= 147.15 m/s

Acceleration,

dv / dt = 9.81

After 15 secs, the acceleration is

9.81 m/s²

-14

step-by-step explanation:

2x2 – 2(x+4)

=2x2-2（5+4）

=4-2×9

=4-18

=-14

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this