Question

A ball was thrown up vertically returns to the thrower after 6 sec . Find - a) the velocity with which it was thrown up . b) the height attained.(g=9.8 m/s^2)

Answer 1

The ball is thrown up vertically...
So.... A=9.8m/s,t=3sec
(time of ascent=time of descent)
V=0
Now... V=u +at
0=u+9.8(3)
u=-29.4
(minus sign indicates that u is in upward direction)
Now.....
For the height attained...
S=ut +1/2at²
S=-29.4(3)+1/2×9.8(3)(3)
S=44.1 m
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Answer 2

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+  1 /2 at2

h=30×3+  1/2 (−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+  1/2 at  2 ′        

where   t  =1 s

′

d=  1/2×10×(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.