Question

a ball thrown upwaeds with velocity 49 time taken to earth
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Answer 1

Answer:

Explanation:

Initial velocity (u) = 49 m/s    Final velocity (v) = 0 m/s

Gravity in toward down = + 9.8 m/s2  Gravity in toward up = - 9.8 m/s2

(i) We know, V2 - u2 = 2gs

             or,  (o)2- (49)2 = 2 X 9.8 X S

             or,  s = - (49)2/ 2 X 9.8 = 122.5 m

Maximum height = 122.5 m

(ii) We know, v = u + gt

             or   0 = 49 +(- 9.8) X t

             or   9.8 X t =  49

             or, t = 49/ 9.8 = 5 s

If, time for upward direction = time for downward direction

Then, total time taken by a ball to return back = 5 + 5 = 10

Answer 2

Answer:

Hi friend

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Your answer

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here,

Final velocity = V = 0 m/s ,

Initial velocity = u = 49 m/s ,

Acceleration due to gravity = a = g = - 9.8 m/s² ,

Height = s = ?

So,

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v² - u² = 2as

=> v² - u² = 2×g×s

=> (0)² - (49)² = 2 × (- 9.8) × s

=> - (49 × 49) = - ( 2 × 9.8 × s)

=> s = (49 × 49) / (2 × 9.8)

=> s = 122.5 m

HOPE IT HELPS