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A ball thrown upward reaches a point 'p' of its path at the end of 4 seconds and the highest point 'q' at the end of 12 seconds. After how many seconds from the start will it reach the point 'p' again
At p when t1=4s
At q when t=12s
At the highest point q,v=0
u=-gt
 = -9.8*12
 = -117.6m/s
distance travelled , S   = u*u/2g
 = (117.6*117.6)/(2*9.8)
 = 705.6m
distance to p from initial = ut-1/2*g*t1^2
    = (-117.6*4) - (1/2 *9.8*4*4)
    = 548.8m
distance between p and q = 705.6-548.8
      = 156.8m
at q, we know v=0
using the equation S= 1/2 *g*t^2
time taken to reach p from q,t2  = sqrt[2S/g]
                = sqrt[2*156.8/9.8]
= 5.6s
so, total time taken = 12+5.6
            = 17.6s


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