Question -
A body of mass m is thrown upwards at an angle θ with the horizontal with speed v. While rising up the speed of the mass after t seconds will be
Options:
A .  √(vcosθ)2+(vsinθ)2
B .  √(vcosθ)2−(vsinθ)2−gt
C .  √v2+g2t2−(2vsinθ)gt
D .  √v2+g2t2−(2vcosθ)gt
Answer: Option C : C Instantaeous speed of rising mass after t sec will be vt = √vx2+vy2 where vx = v sin θ = Horizontal component of velocity vy = v sin θ - gt = Vertical component of velocity vt = √(vcosθ)2+(vsinθ−gt)2
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