Question

If an object is thrown upwards with velocity 3m/s. At what time does it come back to the hand of thrower? Take value of g= 10m/s^2​

Answer 1

u = 3 m/s

g = 10 m/s²

v = 0 m/s [At the highest point, it's velocity is zero]

v = u + at

0 = 3 - 10t

t = 0.3 s

Total time taken (this time includes the path it went through and through which it returned as well) = 0.3 × 2

= 0.6 s

Answer 2

.Here is the best approach to this question.

In the above equation 's' is a displacement vector. By the time the object returns to its initial point of projection, the displacement is zero.

Therefore, s = 0Use the equation,s = ut +½at^2 (taking all vectors vertically upward as positive)0 = 30t + ½x(-10)xt^20 = 6t - t^20 = t(6 - t)Therefore, t = o or 6 sHence, the time taken to return = 6 s