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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

Grade:11

2 Answers

Khimraj
3007 Points
3 years ago
acceleration in horizontal direction is given by qE/m = 5m/s2
time of flight will be given by
0 = 10T – ½gT2
T = 2 sec
maximum height
H = 10*1 + ½*10*12
H = 15m.
Horizontal range = ½*5*22 = 10m.
Hope it clears.
S V Naga Santoshi
13 Points
2 years ago
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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