A small rocket to gather weather data is launched straight up. Several seconds into the flight, its velocity is 140 m/s and it i
1 answer:
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Answer:
B) The 2-kg mass travels twice as far as the 4-kg mass before stopping
Explanation:
As we know that both mass have same horizontal force opposite to their motion
So we will have
now the stopping distance of an object moving with initial speed v is given as
so here we have
for other object we have
So correct answer will be
B) The 2-kg mass travels twice as far as the 4-kg mass before stopping
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.
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.