Answer:
Magnitude of vector A is 51.4 m
The angle of vector A with positive x-axis is θ=
Explanation:
Given is
=-31m
=41m
Let and
are components of a vector A,
The magnitude of vector A will be:
=
A = 51.4 m
we know
θ1=
=
=
Vector A is present in 2nd quadrant as x-component of vector A is negative and y-component is positive.
For finding the angle of vector with positive x-axis,
θ=180-52.9
θ=
Answer:
The new speed must be
Explanation:
In order for the satellite to be on a stable orbit around the planet, the gravitational attraction must be equal to the centripetal force that keeps the satellite in circular motion:
where G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, V0 the speed of the satellite at distance R from the center of the planet.
We can re-write V0, the initial satellite speed, by re-arranging the equation:
Now, if we want the satellite to orbit at a distance of 2R, the new tangential speed must be:
Answer:
a) Maximum height reached = 1878.90 m
b) Time of flight = 39.14 seconds.
Explanation:
Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g and final velocity = 0 m/s.
0 = u sin θ - gt
t = u sin θ/g
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile ,
Vertical motion (Maximum height reached, H) :
We have equation of motion, , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.
Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H
In the give problem we have u = 192 m/s, θ = 90° we need to find H and t.
a)
Maximum height reached = 1878.90 m
b)
Time of flight = 39.14 seconds.