Question

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.

Answer 1

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