Question

A body is thrown vertically upward with speed \( 25 m / s \) at \( t =0 \). The body reaches a certain height at time \( t =2 \) second. It will be at the same height again at the time \( t = t ^{\prime} \) equals to 

(1) 4 seconds 

(2) 3 seconds 

(3) 5 seconds 

(4) \( 2.5 \) seconds

Answer 1

V = u + at

0 = 25 + (-g)+

t = 2.5 sec

h = ut + 1/2 gt²

h = 25 x 2 - 1/2 x 10 x (q)²

h = 50 - 20

h = 30 m

h_max = 25 x 2.5 - 1/2 x 10 x (2.5)²

= 625 - 31.25

h_max = 31.25 m

h_m - h = 1/2 gt²

31.25 - 30 = 1/2 x 10 x (t)²

1.25 = 1/2 x 10 x t²

t = \(\sqrt{\frac{2.5}{10}}\)

t = 0.5 sec

Then 25 + 0.5 = 3 sec