Question

a particle is projected upward with velocity u. find (a) time of ascent (b) time of decent (c) total time of flight (d) maximum height achieved by particle.​

Answer 1

Time of ascent is the time taken to reach the maximum height, after which it starts falling down.

Oppositely, time of decent is time to reach down to the initial position. In absence of air-friction and other forces(except gravity), time of ascent is equal to the time of decent.

At the highest point, it is just stops to fall back. At this point, velocity(v) = 0 and maximum height is reached. Using kinematic eqⁿ v = u + at.

0 = u + (-g)t  → gt = u

→ t = u/g

This is the time to reach max height(time of ascent). Time of ascent = u/g

As time of decent = time of ascent

Time of decent = u/g

Time of flight = time of ascent + time of decent

Time of flight = u/g + u/g

Time of flight = 2u/g

Answer 2

Explanation:

Answer:

20 N

Explanation:

Initially, charges are 6μC and 16μC.

Later on, -8μC is added to both. Then, charges are (6 - 8)μC = -2μC and (16 - 8)μC = 8μC.

q = 6μC, Q = 16μC

q' = -2μC, Q' = 8μC

Force = kqQ/r², let distance be r. Initially:

=> 120 = (9 x 10^9) (6μ*16μ)/r²

=> r² = (9 x 10^9) (96μ²) / 120

=> r² = (864/120) (10^9 x 10^(-6)²)

=> r² = 7.2 x 10^(-3)

When -8μC is added, r is same.Force is:

=> F' = kq'Q'/r²

=> F' = (9 x 10^9) (-2μC*8μC)/(7.2 x 10^(-3))

=> F' = 9 x 10^(9) (-16μ²) / (7.2 x 10^(-3))

=> F' = -(9*6)/7.2 x (10^9 x 10^(-12))/10^(-3)

=> F' = - 20 x 10^0

=> F' = - 20 N

Magnitude of force is 20 N.

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Time of ascent is the time taken to reach the maximum height, after which it starts falling down.

Oppositely, time of decent is time to reach down to the initial position. In absence of air-friction and other forces(except gravity), time of ascent is equal to the time of decent.

At the highest point, it is just stops to fall back. At this point, velocity(v) = 0 and maximum height is reached. Using kinematic eqⁿ v = u + at.

0 = u + (-g)t  → gt = u

→ t = u/g

This is the time to reach max height(time of ascent). Time of ascent = u/g

As time of decent = time of ascent

Time of decent = u/g

Time of flight = time of ascent + time of decent

Time of flight = u/g + u/g

Time of flight = 2u/g

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