Question

The equations of motion of a projectile are given by x =18t, 2y = 54t – 9.8t2
The angle of projection is
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Answer 1

Explanation:

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Answer 2

Answer:

The angle of projection is $tan^-^1(\frac{3}{2})$.

Explanation:

Along horizontal direction

$x=18t$    (1)

Equation (1) is of the form;

$x=ut$     (2)

Where,

x=diaplacement along the horizontal direction

u=velocity along the horizontal direction

t=time taken

By comparing equation (1) with equation (2) we get;

$u=18$      (3)

Along vertical direction

$2y=54t-9.8t^2$

$y=27t-4.9t^2$  (4)

Equation (4) is of the form

$y=vt+\frac{1}{2}at^2$   (5)

y=displacement along the vertical direction

v=velocity along the vertical direction

t=time required to complete the motion

a=acceleration of the particle

By comparing equation (4) with equation (5) we get;

$v=27$     (6)

The angle of projection is given as,

$tan\theta=\frac{v}{u}$    (7)

By substituting the values in equation (7) we get;

$tan\theta=\frac{27}{18}$

$tan\theta=\frac{3}{2}$

$\theta=tan^-^1(\frac{3}{2})$

Hence, the angle of projection is $tan^-^1(\frac{3}{2})$.