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A man in a balloon of 2 m/s relative to the balloon. After what time interval will the ball return to the balloon.
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Let the absolute velocities of the balloon be 'v1' and ball be 'v2'.

  • Given$v_1=5$ and$v_2 - v_1 = 2$
  • Hence$v_2 - 5 = 2$
  • $v_2 = 7$
  • After 't' secs let the ball return to the balloon. Its velocity decreases as the ball moves under gravity, finally ball reaches to the balloon during its upward journey.
  • Displacement of balloon after 't' secs = 5t
  • For ball,$u = 7$ and$a = -g = - 9.81 m/s ^2$.

$s = u t + \frac 1 2 at^2$

$s = 7 t + \frac 1 2 (-9.81)t^2$

$s = 7 t - \frac 1 2 \times 9.81 \times t^2$

$s = 7t - 4.905 \times t^2$........... equation (1)

  • For ball and balloon to meet each other; Displacement of balloon = Displacement of ball
  • Substituting$s = 5t$ in equation (1)
  • Hence$5t = 7t - 4.905 \times t^2$

$4.905 t^2 - 2t = 0$

$t(4.905 t - 2) = 0$

  • $Hence \:t = 0 \,s\: or \:t = 0.4077 \,s$
  • The ball will return to the balloon after a total time interval = 0.4077 s.
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