From the earth's surface, a ball is thrown upward with an initial velocity of v 0 . A drag force equal to m y v 2 affects how the ball moves. (where, m is mass of the ball, v is its instantaneous velocity and gamma is a constant). The ball needs seconds to reach its zenith (maximum height). 1) 1 2 γ g tan − 1 ⁡ ( 2 γ g v o )2) 1 γ g tan − 1 ⁡ ( γ g v o )3) 1 γ g sin − 1 ⁡ ( γ g v o )4) 1 γ g ln ⁡ ( 1 + γ g v o )

Question

From the earth&#039;s surface, a ball is thrown upward with an initial velocity of       v    0  . A drag force equal to   m  y      v    2   affects how the ball moves. (where, m is mass of the ball, v is its instantaneous velocity and gamma is a constant). The ball needs seconds to reach its zenith (maximum height). 1)       1                  2            γ      g            tan          −      1        ⁡  (                    2        γ            g            v    o    )2)       1          γ      g            tan          −      1        ⁡  (            γ      g            v    o    )3)       1          γ      g            sin          −      1        ⁡  (            γ      g            v    o    )4)       1          γ      g        ln  ⁡  (  1  +            γ      g            v    o    )

Raiden Wolf · Accepted Answer

The correct option is 2)       1          γ      g            tan          −      1        ⁡  (            γ      g            v    o    )Given, drag force,   F  =  m  y      v    2    .  .  .  (  i  )Using Newton&#039;s second law,  F  =  m      a    2    .  .  .  (  i  i  )Comparing Eqs. (i) and (ii), we get  a  =  γ      v    2  When thrown vertically up, the ball&#039;s net retardation is      a          n      e      t        =  −  (  g  +  γ      v    2    )  =            d      v              d      t        ⇒            d      v              (      g      +      γ              v        2            )        =  −  d  t  .  .  .  .  .  (  i  i  i  )By integrating both sides of Eq. (iii) when the ball is thrown upward with velocity       v    o   and reaches its zenith, or maximum height at time, within known bounds   t  =  t  ,  v  =  0  ⇒      ∫                  v        o                    O                  d      v              (      γ              v        2            +      g      )        =      ∫          O              t        −  d  tor       1    γ        ∫                  v        o                    O            1          [      (                        g          γ                            )        2            +              v        2            ]        d  v  =  −      ∫          O              t        d  t  ⇒      1    γ        1                  g        γ              [      tan          −      1        ⁡  (      v                  g        γ              )      ]                  v        o                    O        =  −  t  [  ∵  ∫      1                  x        2            +              a        2              d  x  =      1    a        tan          −      1        ⁡  (      x    a    )  ]  ⇒  −      1          γ      g            tan          −      1        ⁡  (                    γ                    v        o                    g        )  =  −  t  ⇒  t  =      1          γ      g            tan          −      1        ⁡  (                    γ                    v        o                    g        )

From the earth's surface, a ball is thrown upward with an initial velocity of vo.

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