East component: 3.9 m/s
South component: 1.8 m/s
Explanation:
We have to resolve the velocity vector along the east and south axis.
Taking east as positive x-direction and south as positive y-direction, the components of the velocity are given by:
where
v = 4.3 m/s is the magnitude of the velocity
is the angle between the direction of the velocity and of the x-axis
Substituting into the equations, we find:
East component:
South component:
Learn more about vector components:
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AnswerShe uses force and air
Explanation:
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Your answer is C)
a)t=2.78 sec
b)R=835.03 m
c)
Explanation:
Given that
h= 38 m
u=300 m/s
here given that
The finally y=0
So
t=2.78 sec
The horizontal distance,R
R= u x t
R=300 x 2.78
R=835.03 m
The vertical component of velocity before the strike
Answer:
The maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km
Explanation:
Using the equations of motion,
When the rocket is fired from the ground,
u = initial velocity = 0 m/s (since it was initially at rest)
a = 10 m/s²
The engine cuts off at y = 0.5 km = 500 m
The velocity at that point = v
v² = u² + 2ay
v² = 0² + 2(10)(500) = 10000
v = 100 m/s
The velocity at this point is the initial velocity for the next phase of the motion
u = 100 m/s
v = final velocity = 0 m/s (at maximum height, velocity = 0)
y = vertical distance travelled after the engine shuts off beyond 0.5 km = ?
g = acceleration due to gravity = - 9.8 m/s²
v² = u² + 2gy
0 = 100² + 2(-9.8)(y)
- 19.6 y = - 10000
y = 510.2 m = 0.510 km
So, the maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km
Hope this helps!!!
