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an arrow is shot into air. its range is 200m and its time of flight is 5s . if g=10m/s2 ,then horizontl compoent of velocity and the mximum height will be

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1. R = Xo * T = 200 m.
Xo*5 = 200
Xo = 40 m/s. = Hor. component of initial
velocity.

2. Tr = T/2 = 5 / 2 = 2.5 s.

Y = Yo + gt.
Yo = Y - gt = 0 - (-10)*2.5 = 25 m/s. =
Ver. component of initial velocity.

hmax = (Y^2-Yo^2)/2g.
hmax = (0-(25)^2) / -19.6 = 31.9 m.

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