A particle is uncharged and is thrown vertically upward from groundlevel with a

Question

A particle is uncharged and is thrown vertically upward from groundlevel with a speed of 25.3 m/s. As aresult, it attains a maximum height h. The particle isthen given a positive charge +q and reaches the samemaximum height h when thrown vertically upward with aspeed of 30.2 m/s. The electric potentialat the height h exceeds the electric potential at groundlevel. Finally, the particle is given a negative chargeq. Ignoring air resistance, determine the speedwith which the negatively charged particle must be thrownvertically upward, so that it attains exactly the maximum heighth. In all three situations, be sure to include the effectof gravity.

Explanation / Answer

   v2   =   u2   +   2* A * h    In firstcase      v   =   0,   u   =   25.3   m/s,   A   =   -9.8   m/s2    02   =   25.32   -2 * 9.8 *h   =>   h   =   640.09/19.6   =   32.66   m    02   =   25.32   -2 * 9.8 *h   =>   h   =   640.09/19.6   =   32.66   m    In second case finalvelocity   v   =   0,   u   =   30.2   m/s,   A   =   effectiveacceleration    0   =     30.22   +   2* A * 32.66    =>   A   =   -912.04 /65.32   =   13.96   m/s2    Acceleration due to electricfield   a   =   A   -g   =   - 13.96 - ( -9.8)   =   -4.16   m/s2    When the particle is given - ve charge, itwill experience same magnitude of acceleration but in oppositedirection. Hence effective acceleration will be    A'   =   - g +a   =   - 9.8 +4.16   =  - 5.64   m/s2    then      v2   =   u2  +   2* A' * h    02   =   u2   +   2* ( - 5.64) * 32.66    required initialvelocity   u   =   368.40   =   19.19   m/s    then      v2   =   u2  +   2* A' * h    02   =   u2   +   2* ( - 5.64) * 32.66    required initialvelocity   u   =   368.40   =   19.19   m/s

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