Mon October 01, 2012 By: Anaida Ann Rose

A particle is projected from the ground in vertical direction at t=0 sec.At t=0.8 sec,it reaches h=14m.It will again come to same height at t= ?

Expert Reply
Mon October 01, 2012
acceleration actiong on the particle: g = -10 m/ss t1 = 0.8 s H = 14 m let initial velocity be u thus using the second equation of motion: H = ut + 1/2gt^2 14 = ut - 5t^2 t1 and t2 be the two solutions thus t1 = 0.8 s and t2 = ? we can solve the equation by first putting in the value of t1 and find u and then find t2.
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