1.

(4) 25 km/h A body is projected vertically upward with speed u. The distance travelled by body in the last second of upward journey is same as that of (1) Distance travelled in first second of downward journey 2) Distance travelled in last second of downward journey (3) Distance travelled in first second of upward journey (4) Both (1) and (2)​

Answer»

Answer:

A body is projected vertically upward with SPEED 40 m/s. The distance travelled by body in the last second of upward journey is [ take G=9.8 m/s  

2  and neglect effect of air resistance]

4.9m

LET the horizontal distance be H and  

let the distance traveled in the last second be d

H=  

2g

u  

2

 

​

 

H=  

19.6

1600

​

 

H=81.63m

Now time taken by the body to REACH a height of 81.36 m=  

g

2H

​

 

​

=4.08s

Now from equation of motion, we have

s=ut+  

2

1

​

GT  

2

 

H−d=(40)(t−1)−  

2

1

​

g(t−1)  

2

 

81.63−d=(40)(4.08−1)−  

2

1

​

(9.8)(4.08−1)  

2

 

d=4.85m

So, the correct answer is '4.9m'.

Explanation:


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