A particle is projected from ground at an angle 45° with initial velocity 20√2ms-1. The magnitude of average velocity in a time interval from t = 0 to t = 3s in ms-1 is.

Question

A particle is projected from ground at an angle 45° with initial velocity 20√2ms-1. The magnitude of average velocity in a time interval from t = 0 to t = 3s in ms-1 is. (A) 20 .62 (B) 10.31 (C) 41 .14 (D) 5.15

Tardigrade Admin · Accepted Answer

u = ux i + uy j uf = uxi +(uy - gt) j, Vavg = (ui+ u f/2)

Q. A particle is projected from ground at an angle 45 $^{\circ}$ with initial velocity $202 m s^{- 1}$ . The magnitude of average velocity in a time interval from t = 0 to t = 3s in $m s^{- 1}$ is.

20 .62

10.31

41 .14

5.15

u = u $_{x}$ i + u $_{y}$ j
u $_{f}$ = u $_{x}$ i +(u $_{y}$ - gt) j, $V_{a vg} = \frac{u _{i} + u _{f}}{2}$

Q. A particle is projected from ground at an angle 45∘ with initial velocity 202​ms−1. The magnitude of average velocity in a time interval from t = 0 to t = 3s in ms−1 is.