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Q.

An object is projected so that it just clears two walls of height 7.5 m and with separation 50 m from each other. If the time of passing between the walls is 2.5 s, the range of the projectile will be: (g = 10 m/s2)

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a

35m

b

70m

c

140 m

d

57.5 m

answer is B.

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Detailed Solution

vx=ux=502.5=20m/s2vyg=2.5⇒vy=12.5m/sNow using, vy2=uy2−2ghWe have vy = 17.5 m/sSo, range =2uxvyg2(20)(1.75)10=70m
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