Question

A stone is projected with a velocity of 5m/s making an angle 50 degrees with the vertical. Is there any other angle of projection the x-y plane having same range? Calculate the range of the projectile. Draw the trajectory of the projectile. What happens time of flight and horizontal range of the is performed in moon?​

Answer 1

Explanation:

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Answer 2

Answer:

The horizontal range of the projectile would also be greater because the projectile would experience less vertical acceleration due to the weaker gravity, allowing it to travel further horizontally before hitting the ground.

Explanation:

From the above question,

  To discover the some other attitude of projection in the x-y airplane which will supply the equal range. The components for the vary R of a projectile is:

                  R = ($v^2$/g) * sin(2θ)

where v is the preliminary velocity of the projectile, θ is the perspective of projection, and g is the acceleration due to gravity.

Given that the preliminary speed of the stone is 5 m/s and the attitude of projection is 50 degrees, we can calculate the vary as:

       R = ($5^2$/9.8) * sin(100 degrees)

       R = 2.5 * sin(100 degrees)

       R ≈ 2.45 m

To locate any other attitude of projection that will provide the equal range, we can use the reality that the sine characteristic has a length of a hundred and eighty degrees.

So, the attitude that will supply the identical vary as 50 tiers is:

       50 + one hundred eighty = 230 degrees

To calculate the trajectory of the projectile, we want to spoil down the preliminary pace into its horizontal and vertical components. The horizontal issue is given by:

        vx = v * cos(θ)

where θ is the attitude of projection and v is the preliminary speed. The vertical factor is given by:

       vy = v * sin(θ)

where θ is the perspective of projection and v is the preliminary speed.

Using these formulas, we can discover that the horizontal issue of the preliminary speed is:

       vx = 5 * cos(50 degrees) ≈ 3.82 m/s

And the vertical issue of the preliminary pace is:

       v_y = 5 * sin(50 degrees) ≈ 3.24 m/s

To discover the trajectory of the projectile, we can use the equations of motion:

       x = v_x * t

       y = v_y * t - (1/2) * g * t^2

g  = 9.8 m/$s^2$.

where x and y are the horizontal and vertical distances traveled by using the projectile at time t, v_x and v_y are the horizontal and vertical factors of the preliminary velocity, and g is the acceleration due to gravity.

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