On planet Tehar, the free-fall acceleration is the same asthat on Earth, but the

Question

On planet Tehar, the free-fall acceleration is the same asthat on Earth, but there is also a strong downward electric fieldthat is uniform near the planet's surface. A 2.0 Kg ball with acharge of 5.0 C is thrown upward at a speed of 20.1 m/s. Ithits the ground after an interval of 4.1 sec. What is the potentialdifference between the starting point and the top point of thetrajectory? NOTE: The Cramster solution is not even close to the correctanswer, so only answer if you really know your Physics. Not a textproblem. I know that youn can use kinematics and V= QLD tosolve. On planet Tehar, the free-fall acceleration is the same asthat on Earth, but there is also a strong downward electric fieldthat is uniform near the planet's surface. A 2.0 Kg ball with acharge of 5.0 C is thrown upward at a speed of 20.1 m/s. Ithits the ground after an interval of 4.1 sec. What is the potentialdifference between the starting point and the top point of thetrajectory? NOTE: The Cramster solution is not even close to the correctanswer, so only answer if you really know your Physics. Not a textproblem. I know that youn can use kinematics and V= QLD tosolve.

Explanation / Answer

   mass m = 2.00 kg    charge q = 5.00 µC    upward speed u = 20.1 m/s    time interval (t) taken to hit theground = 4.10s    the free fall acceleration (g) on earth is sameon tehar so it will be = 9.8 m/s2    when the charge goes up and falls down the netdistance covered will be zero so on applying    the equation of motion we get    0 = ut + (1/2)ayt2    ay = -2u/t    in the upward of the through, let the distancecovered be (s) and the final velocity (v) atmaximum    height will be zero    using the equation of motion    v2 - u2 =2ays we get    s = 0-u2 / 2ay      = -u2 / 2ay    on tehar there are two forces whichare acting downward (negative direction)    one is gravitational force (mg), and other iselectric force (qE)      according to newtons second law of motion    ay = F/m        = -mg-qE / m        = -2u / t    solve the above for E    now the potential difference (V) between thestarting point and the top point of the trajectory    will be    V = (E)(s)        =..........volts    on tehar there are two forces whichare acting downward (negative direction)    one is gravitational force (mg), and other iselectric force (qE)      according to newtons second law of motion    ay = F/m        = -mg-qE / m        = -2u / t    solve the above for E    now the potential difference (V) between thestarting point and the top point of the trajectory    will be    V = (E)(s)        =..........volts

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