Question

hello
please explain the attached Q

good day​

Answer 1

Answer:

MARK IT AS BRAINLIEST HOPE IT HELPS

Answer 2

Hello Mate,

Here is your answer,

Horizontal component =>

$u \cos \alpha$

Vertical component =>

$u \sin \alpha$

The horizontal component remains constant throughout the motion.

Given that at 3 secs in total, the body will travel horizontally

(Which means it has reached its maximum height and its final velocity is zero after 3 secs)

We Know,

$v = u - gt \\ \\ (a = - g) \\ \\ we \: know \: v = 0 \: at \: maximum \: \\ height \\ \\ and \\ \\ v_{y} = u \sin \alpha \\ \\ so \\ \\ \: 0 = u \sin \alpha - (10)(3) \\ \\ u \sin \alpha = 30</p><p></p><p></p><p>when t = 2secs,</p><p></p><p></p><p>[tex]u \cos30 \\ \\ = 20 \times \frac{ \sqrt{3} }{2} \\ \\ = 10 \sqrt{3}$

We Know,

u = square root of Ux^2 + Uy^2

u = 20root3 m/s

TanA = Ux/Uy

TanA = 30/10root3

TanA = 3/root3

TanA = root3

=> A = 60°

Hence Answer is Option: (D)

Hope it helps. :)