Answer:
His speed as he left the floor is 4.83 m/s.
Step-by-step explanation:
Given: 46 inches = 1.1684 m and mass = 200 lb = 90.7185 Kg.
From the third equation of motion under free fall,
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] - 2gs
Where; V is the final velocity (0), U is the initial velocity (unknown), g is the value of gravity - 10 m/[tex]s^{2}[/tex] and s is the distance = 1.1684 m.
Then;
0 = [tex]U^{2}[/tex] - 2gs
[tex]U^{2}[/tex] = 2gs
= 2 × 10 × 1.1684
= 23.368
⇒ U = [tex]\sqrt{23.368}[/tex]
= 4.8340 m/s
The initial velocity, U = 4.83 m/s.
Therefore, his speed as he left the floor is 4.83 m/s.
Answer:
His speed as he left the floor is 4.83 m/s.
Step-by-step explanation:
