A tennis ball served horizontally from a height of 1.8m strikes the ground at an point 18m away from the server.If ot just touches net 12m from the server then height of net is ?
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Height of ball from server = AB = 1.8m
Let the net's height be DE
Let C be the point where the ball strikes the ground, such that
BC=18m
The net is 12m from the server
Hence, EC = 18 - 12 = 6m
So we have two triangles,ABC and DEC
In Triangles ABC and DEC
Angle ACB = Angle DCE (Common angle)
Angle ABC = Angle DEC = 90°
Therefore,
Triangle ABC ~ (is similar to) Triangle DEC (By Angle Angle Axiom)
Therefore,
AB:BC = DE : EC
1.8 / 18 = DE / 6
1.8 / 18 * 6 = DE
DE = 0.6m
Hence, Height of net = 0.6m(Ans)
Let the net's height be DE
Let C be the point where the ball strikes the ground, such that
BC=18m
The net is 12m from the server
Hence, EC = 18 - 12 = 6m
So we have two triangles,ABC and DEC
In Triangles ABC and DEC
Angle ACB = Angle DCE (Common angle)
Angle ABC = Angle DEC = 90°
Therefore,
Triangle ABC ~ (is similar to) Triangle DEC (By Angle Angle Axiom)
Therefore,
AB:BC = DE : EC
1.8 / 18 = DE / 6
1.8 / 18 * 6 = DE
DE = 0.6m
Hence, Height of net = 0.6m(Ans)
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