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A body is projected vertically upward with a velocity of 40 m/s from the foot of a tower. If it crosses the mid point of the tower twice in a time interval of 4 s, height of the tower is

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Detailed Solution

when the body is passing the mid point of the tower, let its velocity be ‘V' m/s∴ 2Vg=4⇒V=4×102 m/s=20 m/sNow, V2=u2−2.g.H2⇒H=402−20210m=120 m

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