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
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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:
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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