Question

A particle is projected such that st t = 3sec and t=5 sec particle is at same height . if speed of particle at maximum height is 40 m/s .​

Answer 1

Answer:

By first equation the maximum height reached

v=u+at

0=40+(−g)(t)

t=

g

40

=4sec

Then, the drop of particle from maximum height

s=ut+

2

1

at

2

[6sec=4sec+2sec]

s=

2

1

(g)(2)

2

=20m

Maximum height reached by particle

s=ut+

2

1

at

2

s=40(4)+

2

1

(10)(4)

2

=40×4+5×16

s=160+80=240m

Thus,

Displacement =240−20=220m

Distance =240+20=260m