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A projectile is projected from top of tower with velocity 50 m/s at $37^{0}$ above with horizontal. Velocity of the particle after 1 sec $(C o s 53^{0} = \frac{4}{5}) (g = 10 m / \sec^{2})$

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10 m/sec

44m/sec

20 m/sec

20.55 m/sec

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detailed solution

Correct option is B

given initial velocity U=50 m/s; θ=370; time =t=1 sec;g=10m/s2Ux=U cosθUx=50×45Ux=40m/secVx=Ux+at ; here a=0 =horizontal accelerationVx=40m/secUy=UsinθUy=50×35=30m/sVy=Uy−gtVy=30−10(1)=20m/secV=Vx2+Vy2=402+202=1600+400=44.72m/sec

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A projectile is projected from top of tower with velocity 50 m/s at 370 above with horizontal. Velocity of the particle after 1 sec Cos 530=45 g=10m/sec2

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A projectile is projected from top of tower with velocity 50 m/s at $37^{0}$ above with horizontal. Velocity of the particle after 1 sec $(C o s 53^{0} = \frac{4}{5}) (g = 10 m / \sec^{2})$