Answer:
the jumper's speed just as he leaves the ground = 4.2 m/s
the force he must exert on the ground to perform the 0.90 m jump = 4.99 × 10³ N
Explanation:
Given that :
height h = 0.90 m
mass = 51 kg
distance s = 0.10 m
What is the jumper's speed just as he leaves the ground?
The velocity P.E = K.E
[tex]mgh = \frac{1}{2} mv^2[/tex]
[tex]2gh = v^2[/tex]
[tex]v = \sqrt{2gh}[/tex]
where;
g = 9.8
[tex]v = \sqrt{2* 9.8*0.9}[/tex]
v = 4.2 m/s
the jumper's speed just as he leaves the ground = 4.2 m/s
What force must he exert on the ground to perform the 0.90 m jump?
Let's first determine the acceleration by using the equation of motion;
v² = u² + 2as
4.2² = 0 + 2(a)(0.10)
17.64 = 0.2 a
a = 17.64/0.2
a = 88.2 m/s²
The force F is now calculated by the relation:
F - mg = ma
F = mg+ ma
F = m(g+a)
F = 51(9.8 + 88.2)
F = 4988 N
F = 4.99 × 10³ N
