Respuesta :
Answer:
a) 50.4 m/s
b) 122.5 m
c) 47.61 m/s , 122.5 m
Explanation:
a)
v₀ = initial velocity of the small mailbag = - 1.39 m/s
a = acceleration = - 9.8 m/s²
v = final velocity after "5 sec"
t = time interval = 5 sec
using the equation
v = v₀ + a t
v = - 1.39 + (- 9.8) (5)
v = - 50.4 m/s
speed of mailbag after "5 sec" is 50.4 m/s
b)
D = distance traveled by helicopter in "5 sec "
d = distance traveled by small mailbag in "5 sec"
distance traveled by helicopter in "5 sec " is given as
D = v₀ t
D = (- 1.39) (5)
D = - 6.95 m
distance traveled by small mailbag in "5 sec " is given as
d = v₀ t + (0.5) a t²
d = (- 1.39) (5) + (0.5) (- 9.8) (5)²
d = - 129.45 m
y = distance below the helicopter
distance below helicopter is given as
y = |d - D| = |(- 129.45) - (- 6.95)| = 122.5 m
c)
v₀ = initial velocity of the small mailbag = 1.39 m/s
a = acceleration = - 9.8 m/s²
v = final velocity after "5 sec"
t = time interval = 5 sec
using the equation
v = v₀ + a t
v = 1.39 + (- 9.8) (5)
v = - 47.61 m/s
speed of mailbag after "5 sec" is 47.61 m/s
D = distance traveled by helicopter in "5 sec "
d = distance traveled by small mailbag in "5 sec"
distance traveled by helicopter in "5 sec " is given as
D = v₀ t
D = (1.39) (5)
D = 6.95 m
distance traveled by small mailbag in "5 sec " is given as
d = v₀ t + (0.5) a t²
d = (1.39) (5) + (0.5) (- 9.8) (5)²
d = - 115.55 m
y = distance below the helicopter
distance below helicopter is given as
y = |d - D| = |(- 115.55) - (6.95)| = 122.5 m
