Question

A ball thrown up vertically returns to the thrower thrower after 9 second. a) find the velocity with which it was thrown up. b) the maximum height it reaches c) and its position after 5.5 second​

Answer 1

Answer:

Explanation:

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.