kinematics question jee

Question

A particle is projected vertically from the ground takes time t_{1} = 1 s upto point A, t_{3} = 3 s from point A to B, and time t_{3} = 4 s from point B to highest point. Find the height of the middle point of A and B from the ground.

Answer 1

let g = 10m/s^2

Total time taken to reach the Highest point = 1 + 3 + 4 = 8 sec

at Highest point final velocity will be 0
then v = u - gt

0 = u - 10(8)

then u = 80 m/s

Hight of point A = distance travelled in 1

A = ut - 1/2gt^2 ( substitute u = 80 and t =1 )

A = 80 - 1/2(10)(1) = 75 meters

Hight of point B = distance travelled in 4

B = ut - 1/2gt^2 ( substitute u = 80 and t =4 )

B = 80(4) -1/2(10)4^2 = 320 - 80 = 260 meters

Mid point of A and B = (75+260)/2  = 167.5 meters