Question

A ball thrown up vertically returns to the thrower after 6 s (given g=10m/s2). Find: (a) the velocity with which it was thrown up, (b) the maximum height it reaches, and (c) its position after 4s.​

Answer 1

Answer:

(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey. Hence, it has taken 3 s to attain the maximum height. Hence, the ball was thrown upwards with a velocity of 29.4 m s−1.

Explanation:





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Class 11

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Question



A ball thrown up vertically returns to the thrower after 6s. Find

(a) The velocity with which it was thrown up,

(b) The maximum height it reaches, and

(c) Its position after 4s.

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Solution



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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+21at2

h=30×3+21(−10)×32              

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+21at′2                        where   t′=1 s

d=21×10×(1)2    =5 m  

∴ Its height above the ground, h′