the equation of projectile is y=16x-(5x*5x)/4.the horizontal range is ?
A.16m
B.8m
C.3.2m
D.12.8m
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for y=0
x=64/25,0
×=2.56m
slope of projectile=dy/dxatx=0=16-50x/4=16
x=64/25,0
×=2.56m
slope of projectile=dy/dxatx=0=16-50x/4=16
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