Question

A particle is projected from ground at an angle of 45° with initial velocity 20 root 2m/s.
The magnitude of average velocity in a time interval from t=0 to t = 3 s in ms 'is​

Answer 1

Answer:

$v = \sqrt{(-10)^2 +20^2}= \sqrt{500} = 10\sqrt{5} m/s$

hope it helps, please rate

Explanation:

vx= 20 sqrt 2. cos45 = 20sqrt2 . 1/2 sqrt2 = 20.2/2 = 20m/s

vy=20 sqrt 2. sin45 = 20sqrt2 . 1/2 sqrt2 = 20.2/2 = 20m/s

vx will be constant only vy will not because of gravitation effect

then, assume the particle is moving upward

assume g =10/m^2

vy_final = vy_initial + g .t

vy_final = 20-10.3 = -10m/s( - means downward

then

vy average = (vy_final - vy_inital) / TimeInterval

vy average = (-10 - 20 )/3s = -10m/s

then the magnitude will be

$v = \sqrt{(-10)^2 +20^2}= \sqrt{500} = 10\sqrt{5} m/s$