At t = 0, a particle leaves the origin with a velocity of 5.0 m/s in the positive y direction. its acceleration is given by = (3.0 - 2.0) m/s2. at the instant the particle reaches its maximum y coordinate how far is the particle from the origin?

The distance of the particle from the origin after reaching its maximum is; 11.267m

We are given;

Initial velocity; u = 5j (because it is in the positive y direction)

Acceleration; a = 3i - 2j

At maximum point, final velocity is zero. Using newton's first equation of motion in the y direction gives;

v = u + at

0 = 5j + t(-2j)

t = 5j/2j

t = 2.5 secs

Using newtons third equation of motion, we can find the distance in the y-direction as;

y = ut + ¹/₂at²

y = 5(2) + (¹/₂ * -2 * 2.5²)

y = 6.25 m

Distance in the x-direction is;

x = (0 * 2.5) + ¹/₂ * 3 * 2.5²

x = 9.375 m

Thus, distance from the origin is;

d = √(x² + y²)

d = √(6.25² + 9.375²)

d = 11.267m

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d = 11.27 m

Explanation:

initial velocity of the particles at origin is given as

acceleration is given as

now when particle will reach to maximum Y coordinate then its velocity in Y direction becomes zero

t = 2.5 s

now we have

Similarly for x coordinate we have

Now the distance from origin is given as

b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

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c

explanation: