Answer:
(C) Average Velocity = 25 m/s due North
Answer:
a) F = 1.26 10⁵ N, b) F = 2.44 10³ N, c) F_net = 1.82 10³ N directed vertically upwards
Explanation:
For this exercise we must use the relationship between momentum and momentum
I = Δp
F t = p_f -p₀
a) It asks to find the force
as the man stops the final velocity is zero
F = 0 - p₀ / t
the speed is directed downwards which is why it is negative, therefore the result is positive
F = m v₀ / t
F = 63.5 7.89 / 3.99 10⁻³
F = 1.26 10⁵ N
b) in this case flex the knees giving a time of t = 0.205 s
F = 63.5 7.89 / 0.205
F = 2.44 10³ N
c) The net force is
F_net = Sum F
F_net = F - W
F_net = F - mg
let's calculate
F_net = 2.44 10³ - 63.5 9.8
F_net = 1.82 10³ N
since it is positive it is directed vertically upwards
Converting 209 km/hr to m/s:
209 km/hr = 58.0 m/s
Contact distance = 0.1 m
Using the equation of motion:
2as = v² - u², where initial velocity is 0
a = (58)² / (2 x 0.1)
a = 1.68 x 10⁴ m/s²
Answer:
the answer is B. it's too easy
