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= ?
Asked by Anaida Ann Rose | 1st Oct, 2012, 11:01: PM
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.
Answered by | 1st Oct, 2012, 11:16: PM
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