Refer to the diagram shown below.
u = 0, the initial vertical velocity
Assume g = 9.8 m/s² and ignore air resistance.
At the first stage of landing on the ground, the distance traveled is
h = 3.1 - 0.6 = 2.5 m.
If v = the vertical velocity at this stage, then
v² = u² + 2gh
v² = 2*(9.8 m/s²)*(2.5 m) = 49 (m/s)²
v = 7 m/s
At the second stage of landing on the ground, let a = the acceleration (actually deceleration) that his body provides to come to rest.
The distance traveled is 0.6 m.
Therefore
0 = (7 m/s)² + 2(a m/s²)*(0.6 m)
a = - 49/1.2 = - 40.833 m/s²
Answers:
(a) The velocity when the man first touches the ground is 7.0 m/s.
(b) The acceleration is -40.83 m/s² (deceleration of 40.83 m/s²) to come to rest within 0.6 m.
Explanation:
Covalent or molecular compounds form when elements share electrons in a covalent bond to form molecules. Molecular compounds are electrically neutral. Ionic compounds are (usually) formed when a metal reacts with a nonmetal (or a polyatomic ion). Covalent compounds are formed when two nonmetals react with each other.
Acceleration = Change in Velocity / time
a = (v - u) / t
Where v = final velocity in m/s
u = initial velocity in m/s
t = time in seconds.
a = acceleration in m/s²
A proper record of the changes in velocity with the corresponding time would help find the acceleration.