A projectile is thrown with a velocity of 10√2m/s at an angle of 45∘ with horizontal. The interval between the moments when speed is √125m/s is (g=10m/s2)
A
1s
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B
1.5s
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C
2s
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D
0.5s
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Solution
The correct option is A1s ux=10√2cos45∘=10m/suy=10√2sin45∘=10m/s Let α be the angle with the horizontal when speed is =√125m/s ; As the horizontal component of velocity always remains constant 10√2cos45∘=√125cosα⇒cosα=2√5 i.e.vy=√125×sinα=5√5×1√5=5m/s
Using first equation of motion: vy=uy−g(t1)⇒5=10−10t1⇒t1=12s During downward journey: −vy=uy−g(t2)⇒−5=10–10t2⇒t2=32s ∴t2−t1=1s The interval between the moments when speed is √125m/s is 1s.
Alternative solution
Horizontal component will remain constant 10√2cos45∘=√125cosα⇒cosα=2√5⇒sinα=1√5 For distance AB in path of projectile, Time period T=2vsinαg=2√125×1√510=1s