Q1. A stone is thrown vertically upwards and its reaches back to ground after 5 seconds find u and the max height stone will attain.
Q1. A stone is thrown vertically upwards and its reaches back to ground after 5 seconds find 'u' and the max height stone will attain. Solution: s = ut + 1/2 at^2 0 = u(5) + 1/2(-9.81)(5)^2 (9.81/2)(5×5/5) = u u = 4.905×5 u= 24.55 m/sec Now we find maxi height V^2 = u^2 + 2as 0 = (24.55)^2 + 2(-9.81)s 24.55×24.55/2(9.81) = s s = 30.72 m