Question

A ball is thrawn vertically returns to the thrower after 10 s. What is the position of the object after 8s

plz give me answer fast plz​

Answer 1

Answer:

Solution

The ball returns to the ground after 6 seconds.

Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s

Let the velocity with which it is thrown up be  u

(a). For upward motion,                

v=u+at

∴     0=u+(−10)×3                  

⟹u=30  m/s      

(b). The maximum height reached by the ball

h=ut+  

2

1

​

at  

2

 

h=30×3+  

2

1

​

(−10)×3  

2

               

h=45 m  

(c). After 3 second, it starts to fall down.  

Let the distance by which it fall in 1 s   be   d

d=0+  

2

1

​

at  

′2

                        where   t  

′

=1 s

d=  

2

1

​

×10×(1)  

2

    =5 m  

∴ Its height above the ground, h  

′

=45−5=40 m  

Hence after 4 s, the ball is at a height of 40 m above the ground.  

Explanation:

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Answer 2

Answer:

please tell after height is 40 second