Question

Question a ball thrown up vertically returns to the thrower after 6 seconds find the velocity with which it was thrown up the maximum height it reaches and the position

Answer 1

Answer:

after analysis of question

u=?

v=0

t=6/2=3 sec

a=-9.8m/s2

therefore

v=u+at

0=u+(-9.8)(3)

u=29.4m/s

the velocity with which it was thrown =29.4 m/s

the distance it travelled

s=ut + a(tsquare)/2

=29.4*3+(-9.8)(9)/2

=88.9+44.1

=187.3 m

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.