Question

At a height 0.4 m from the ground the velocity of particle projected in angle θ is ˆv=(6i+2j)m/s The angle of projection θ is (g=10m/s2)

Answer 1

Given, v = 6i + 2j
means, x component of velocity is 6 e.g., $v_x$ = 6
y component of velocity is 2 e.g., $v_y$ = 2
We know, in case of projectile motion, horizontal component velocity remains constant. So, $v_x = u_x$ = 6

And for y - component, $v_y^2=u_y^2+2a_yh$
2² = $u_y^2$ - 2 × 10 × 0.4
$u_y^2$ = 12
so, $u_y$ = √12 = 2√3

Now, $Tan\theta=\frac{u_y}{u_x}$
= 2√3/6 = 1/√3 = tan30°

Hence, θ = 30°

Answer 2

Answer:

Explanation:Given, v = 6i + 2j

means, x component of velocity is 6 e.g., = 6

y component of velocity is 2 e.g., = 2

We know, in case of projectile motion, horizontal component velocity remains constant. So, = 6

And for y - component,

2² = - 2 × 10 × 0.4

= 12

so, = √12 = 2√3

Now,

= 2√3/6 = 1/√3 = tan30°

Hence, θ = 30°

Read more on Brainly.in - https://brainly.in/question/4378224#readmore