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```
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

```
5 years ago

Hasan Naqvi
97 Points
```							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
```
5 years ago
Pavan
13 Points
```							Why this question u is take 0 n in question given that body is projected vertically upward with speed 40m/s
```
3 years ago
Ashish Sahani
21 Points
```							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.
```
3 years ago
ankit singh
596 Points
```							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.thanks and regards
```
5 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions