A particle is projected from ground at angle 45^@ with initial velocity 20(sqrt2) m//s. Find (a)change in velocity, (b)magnitude of average velocity in a time interval from t=0 to t=3s.
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Angle of the particle = 45° (Given)
Initial Velocity = 20√2 m/s (Given)
a) As acceleration is constant, thus -
a = g = ( - 10j) = ΔvΔ/t
Δv = ( - 10j) (Δt) = ( - 10j) (3)
= ( - 10j) (3)
= - 30j m/s
Therefore, the change im velocity is 30m/s vertically downwards
b) v = s/t
= ut + 1/2 at² / t
= u + 1/2 at
= ( 20i + 20j) + 1/2 (-10j) (3)
= ( 20i + 20j) - 15j
= 20i + 5j
Therefore,
V = √20² + 5²
= √ 400 + 25
= 20.6 m/s
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