if the angle of projection of a particle from the horizontal is doubled keeping the speed of projection same, the particle strikes the same target on the ground, then the ratio of time of flight in the two cases will be :
Answers
1: √3 will be ratio of time of flight in 1st to 2nd case
Step-by-step explanation:
Let say speed = V
Horizontal Speed = Vcosα
Vertical speed = VSinα
Time to reach peak = VSinα/g
Total time of flight = 2VSinα/g
Horizontal Distance = Vcosα * 2VSinα/g = V²Sin2α /g
when the angle of projection of a particle from the horizontal is doubled
Horizontal Speed = Vcos2α
Vertical speed = VSin2α
Time to reach peak = VSin2α/g
Total time of flight = 2VSin2α/g
Horizontal Distance = Vcos2α * 2VSin2α/g
Vcos2α * 2VSin2α/g = V²Sin2α /g
=> cos2α * 2 = 1
=> cos2α = 1/2
=> 2α = 60°
=> α = 30°
Total time of flight in 1st case = 2VSinα/g = 2VSin30/g
Total time of flight in 2nd case = 2VSin2α/g = 2VSin60/g
Ratio 1st/2nd = Sin30/Sin60 = 1/√3
1: √3 will be ratio of time of flight in 1st to 2nd case
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