On planet Tehar, the free fall acceleration is the same as that of Earth, but there is also a strong downward electric field that is uniform close to the planet's surface. A 4.05 kg ball having a charge of 4.65 [tex] \mu C [/tex] is thrown upward at a speed of 22.2 m/s, and it hits the ground after an interval of 1.86 s. What is the potential difference between the starting point and the top point of the trajectory? Answer in units of kV.(adsbygoogle = window.adsbygoogle || []).push({});

First I drew a free body diagram and found that the only forces on the ball were the force of the charge and gravity.

So qE + mg = ma

where a is constant.

Then I figured out the height of the ball by adding the final and initial velocities and dividing by 2 and multiplying by the time/2.

22.0+ 0 /2 *1.86/2

h= 10.323

Then I used conservation of energy

KE_o + PE_0 +EPE_o = KE_f + PE_f + EPE_f

1/2mv^2_o +0 +0 = 1/2 mv^2_f + mgh +qV

So V= 1/q ((1/2)mv^2_0 -mgh)

V= 1/4.65 x 10^-6 ((1/2)4.05(22.2^2) - 4.05 (9.8)( 10.323))

V=1.26x 10^8

V= 1.26 x 10^4 kV

This isn't right.. can someone tell me what's wrong?

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# Homework Help: Electric potential and free fall acceleration problem

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