A projectile is thrown with a velocity of 10√2ms−1 at an angle of 45o with horizontal. The time interval between the moments when its speed is √125ms−1 is g=10ms−2
A
1.0s
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B
0.6s
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C
1.2s
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D
0.8s
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Solution
The correct option is A1.0s v is the velocity of the particle where the speed is √125ms−1
As the horizontal velocity is constant, ucosθ=vcosψ 10√2×1√2=√125cosψ cosψ=2√5 ⇒sinψ=√1−(25)2=1√5
Between points A and B t=2vsinψg=1s