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Here, the horizontal component of initial velocity will be vcosq and vertical component of velocity will be vsinq.
The horizontal distance travelled in time t will be
x = (vcosq)t – ½ gt2
The vertical distance travelled in time t will be
x = (vsinq)t – ½ gt2
so, we have
x= (vcosq)(2vsinq/g) + ½ a(2vsinq/g)2
and for maximum horizontal range, we differentiate (1) w.r.t q
dx/dq =cos2q+a/g sin2q =0
(or)
Tan2q = –g/a = –g/QE/m = –mg/QE
Thus, the maximum distance the object can travel horizontally before striking ground is
– mg / QE
Thanks & Regards,
Vasantha Sivaraj,
askIITians faculty
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