Question

For a particle x, y coordinates, varies with times
as, X = 6t, y = 8t - 5t2. The initial speed of
projection is (x and y are in meter and t in
second)
(1) 10 m/s
(2) 100 m/s What
(3) 20 m/s
(4) V10 m/s
stion of nroiontila ich​

Answer 1

Explanation:

this. is. your. answer.

Answer 2

Answer:

The initial speed of the projection is 10 m/s.

Explanation:

Given,

x = 6t

y = 8t - 5t²

Where x represents the motion in the x-direction, y represents the motion of the body in the y-direction, and t represents the time.

To find,

The initial velocity of the projection.

Concept,

The velocity of a body at any point in time t is given by (V)

$V = \sqrt{V_x^2(t)+V_y^2(t)}$

Where $V_x$ is velocity along the x-direction and given by the first derivative w.r.t time (t) of the motion along the x-direction, and $V_y$ is velocity along the y-direction and given by the first derivative w.r.t time (t) of the motion along the y-direction.

Calculation,

Along x-direction:

x = 6t

$\frac{dx}{dt} = V_x = 6 m/s$

At t = 0 velocity along x-direction = 6 m/s

Along y-direction:

y = 8t - 5t²

$\frac{dy}{dt} = V_y = 8 - 10t$

At t = 0 velocity along y-direction = 8 m/s

Now, the resultant velocity (V) is

$V = \sqrt{V_x^2(t)+V_y^2(t)}$

$V =\sqrt{6^2+ 8^2}$

V = 10 m/s

Therefore, the initial speed of the projection is 10 m/s.

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