A BODY IS PROJECTED WITH A VELOCITY OF 40 M/S . AFTER 2 SEC IT CROSSES A POLE OF 20.4m . FIND THE ANGLE OF PROJECTION AND RANGE OF HORIZONTAL MOVEMENT
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let inclination angle =∅
so, y -component of initial velocity = 40sin∅
equation for y-component
so, Y = Uyt + 1/2ayt²
20.4 = 40sin∅ ×2 -1/2g(2)²
20.4 =80sin∅ -19.6
40 = 80sin∅
sin∅ = 1/2 = sin30°
∅ = 30° ( answer)
Range = u²sin2∅/g
= (40)² sin60°/g
= 1600×√3/2 ×1/g
=800√3/g m
so, y -component of initial velocity = 40sin∅
equation for y-component
so, Y = Uyt + 1/2ayt²
20.4 = 40sin∅ ×2 -1/2g(2)²
20.4 =80sin∅ -19.6
40 = 80sin∅
sin∅ = 1/2 = sin30°
∅ = 30° ( answer)
Range = u²sin2∅/g
= (40)² sin60°/g
= 1600×√3/2 ×1/g
=800√3/g m
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