A particle of mass 2kg and charge 1mC is projected vertically with a velocity 10m/s. THere is a uniform horizontal electric field of 10^4 N/C. Find the horizontal range, the time of flight and the mazimum height attained by the particle

Question

Khimraj · Answer

acceleration in horizontal direction is given by qE/m = 5m/s 2 time of flight will be given by 0 = 10T – ½gT 2 T = 2 sec maximum height H = 10*1 + ½*10*1 2 H = 15m. Horizontal range = ½*5*2 2 = 10m. Hope it clears.

S V Naga Santoshi · Answer

Acceleration in horizontal direction is qE/m Since F=qE in this case...so by substituting the values a=5m/s^2.. To find time of flight we know that .. S=ut+1/2at^2 i.e, at 0=10T -1/2×10×T^2 then we will get T=2seconds... Maximum height H : V^2-U^2 =2gH -100 = 2× (-10) ×H .. [since body moving upward direction g is negative and at maximum height v=0] H = 5m … Thanking you for giving an opportunity... ifthere is any mistake please forgive me..

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A particle of mass 2kg and charge 1mC is projected vertically with a velocity 10m/s. THere is a uniform horizontal electric field of 10^4 N/C. Find the horizontal range, the time of flight and the mazimum height attained by the particle

A particle of mass 2kg and charge 1mC is projected vertically with a velocity 10m/s. THere is a uniform horizontal electric field of 10^4 N/C. Find the horizontal range, the time of flight and the mazimum height attained by the particle

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