On planet Tehar, the free-fall acceleration is the same as that onEarth, but the

Question

On planet Tehar, the free-fall acceleration is the same as that onEarth, but there is also a strong downward electric field that isuniform close to the planet's surface. A 1.98 kg ball having a charge of 4.96 µC is thrown upward at a speed of 20.1m/s. It hits the ground after an interval of 4.10 s. What is thepotential difference between the starting point and the top pointof the trajectory?
kV

On planet Tehar, the free-fall acceleration is the same as that onEarth, but there is also a strong downward electric field that isuniform close to the planet's surface. A 1.98 kg ball having a charge of 4.96 µC is thrown upward at a speed of 20.1m/s. It hits the ground after an interval of 4.10 s. What is thepotential difference between the starting point and the top pointof the trajectory?
kV

On planet Tehar, the free-fall acceleration is the same as that onEarth, but there is also a strong downward electric field that isuniform close to the planet's surface. A 1.98 kg ball having a charge of 4.96 µC is thrown upward at a speed of 20.1m/s. It hits the ground after an interval of 4.10 s. What is thepotential difference between the starting point and the top pointof the trajectory?
kV

Explanation / Answer

   mass m = 1.98 kg    charge q = 4.96 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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