Answer
given,
distance of fall = h
initial speed = 0 m/s
it travels 0.540 h in the last 1.00 s
Speed of the fall = [tex]\dfrac{Distance}{time}[/tex]
Speed = [tex]\dfrac{0.540\ h-0}{1}[/tex]
S = 0.540h m/s
time of the fall
using equation of motion
v = u + g t
0.540 h = 0 + 9.8 t
t = 0.055h s
The time of fall is equal to 0.055h s
height of fall = ?
again using equation of motion
v² = u² + 2 g s
(0.540h)² = 0 + 2 x 9.8 x s
s = 0.0148 h²
Hence, the time of fall of the object = 0.055h s.
the distance of fall of object = 0.0148 h²
