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Question

A ball thrown vertically upward returns to the thrower after 6 s. The ball is 5 m below the highest point at t = 2 s. The time at which the body will be at same position, (take g = 10m/s2).

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Solution

Total time of flight(t)=6s
So time to reach the maximum height =3s
if u= projected velocity then:
0=ugt2u=gt2=602=30 m/s
Applying equation v2=u22gh(for half journey)
0=90020h
h=45m
When the ball is at height 5m from highest point, it's height from ground =455=40m
Applying equation: s=ut+12at2(at h=40m)
40=30t12gt2
40=30t5t2t26t+8=0
As one root of the equation is given as: t1=2s
By product of roots t1t2=8
t2=4s. (when the ball reaches the same height again)

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