Question

A particle is projected vertically upwards from the
earth. It crosses the same point at t = 2 sec and
t = 8 sec. Find out lg = 10 ms]
(i) the time for which particle remains in air
(ii) initial velocity of particle
(iii) maximum height attained by the particle
(iv) height of the particle at t = 2 sec and t = 8 sec
from earth​

Answer 1

Answer:

10 sec

50 m/s

125 m

80 m

Explanation:

A particle is projected vertically upwards from the earth. It crosses the same point at t = 2 sec and t = 8 sec.

Let say Initial Velocity = V

S = Ut + (1/2)at²

a = g = -10 m/s

Distance in 2 sec

S = 2V + (1/2)(-10)2²

S = 2V - 20

Distance in 8sec

S = 8V + (1/2)(-10)8²

S = 8V - 320

2V - 20 = 8V - 320

=> 6V = 300

=> V = 50 m/s

Initial Velocity = 50 m/s

Time to reach top = 50/10 = 5 sec

Total Time of Flight = 2 *5 = 10 sec

particle remains in air for 10 sec

maximum height attained by the particle

= 50²/(2*10)

= 125 m

S = 2V - 20 = 2 *50 - 20 = 80 m

height of the particle at t = 2 sec and t = 8 sec from earth​  = 80 m

Answer 2

the answer to your question is in the pic