Question

A ball is thrown upwards and return to the thrower at 6s Find velocity it was thrown upwards Maximum height Its position after 4s

Answer 1

total time of journey of ball =T= 6seconds
we know that time of Ascend = time of descend
i.e. the ball will take same time in going up and coming down
=> time to go up= 6/2=3seconds. = time to come down
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CASE 1
(from the instance it is thrown up to the time it reaches maximum height)

initial velocity of throwing = um/s (let)
final velocity= 0m/s {as the body momentarily comes to rest}
time=3seconds
a= -g = -9.8m/s^2 {-ve because it is acting downwards}
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(i) using 1st equation of kinematics:-

=>v=u+at

=> 0=u-(9.8)(3)

=> u=9.8(3) = 29.4m/s
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(ii) using 3rd equation of kinematics:-

s=ut+1/2(a)t^2

=>s=29.4(3)-1/2(9.8)(9)

=>s= 44.1m
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hope it helps you...
☺

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.