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Grade 11Mechanics

A body is projected vertically upward with speed 40m/s. The distance travelled by body in the last second of upward journey is.take g as 9.8

Profile image of Soumyadeep
11 Years agoGrade 11
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4 Answers

Profile image of Hasan Naqvi
11 Years ago
Since the distance travelled in the last second of upward journey = distance travelled in first second of downward journey(the height-time graph is symetric about the highest point)
 
s = ut + ½ at2
= 0 + 0.5 * 9.8 * 1
=4.9m
Profile image of Pavan
8 Years ago
Why this question u is take 0 n in question given that body is projected vertically upward with speed 40m/s
Profile image of Ashish Sahani
8 Years ago
niche aate samay jo pahle second me distance covered karega wahi last second ke upar jane me bhi karega.Isliye niche aane ke pahle V=0m/s ho jayegaphir wo niche aayega to ab U=0m/s lenge aur pahle second me distance travelled nikal lenge jo upar jate samay last second ke distance ke barabar hota hai.
Profile image of ankit singh
5 Years ago
body is projected vertically upward with speed 40m SThe distance by body in the last second of upward journey is [take g=9.8 m/s2 and neglect effect of air resistance ] 1)4.9 m.
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