Question

A particle is projected vertically upwards with a velocity U. After an interval of time t another particle is projected upwards from the same point and with the same initial velocity. Prove that particles will meet at the height (4u-gt)/8g​

Answer 1

Step-by-step explanation:

Vertical POSITION will be found by

Ut - 1/2 a t^2 where a = 9.81 m/s^2

the SECOND particle position will be

U(s+t) - 1/2 a ( s+t)^2

equate them to find when they are at the same position

Us - 1/2 as^2 = Us + Ut - 1/2 a ( s^2 + 2st + t^2 )

0 = Ut - 1/2 a ( 2st+ t^2 )

0 = Ut - ast - 1/2 a t^2