a projectile is fixed from ground level with a velocity 500m/s at 30 degree to the horizontal .find its horizontal range, the greatest height and time to reach the greatest height.

1. 22092.31 m

2. 3188.78 m

3. 25.51 s

Explanation:

Let's split this projectile's motion into its horizontal and vertical components:

Since this projectile is in constant acceleration (9.8 m/s² vertically downwards and 0 m/s² horizontally), we can use the kinematic equations for constant acceleration.

Let's find the time it takes for the projectile to reach its greatest height first by using our knowledge that, at the top of its trajectory, the projectile will have a final velocity of 0 m/s before changing its direction and falling down.

Therefore, we can use the equation that contains the variables a_y, v_f, v_i, and t in order to solve for time t.

This equation is: v = v_0 + at, where v is the final velocity, v_0 is the vertical component of the initial velocity, a is the acceleration in the y-direction, and t is the time of flight of the projectile.

Plug known values into the equation:

Subtract 500 * sin(30) from both sides of the equation.

Divide both sides by -9.8.

The time it takes the projectile to reach its greatest height is 25.51 seconds.

Now we can use this time t to find the greatest height of the object, which will be the displacement in the y-direction at t = 25.51 seconds.

We can use this equation that contains the variables displacement (Δx), initial velocity (v_0), acceleration (a_y), and time (t):

Plug known values into the equation.

The displacement in the y-direction is 3188.78 m at half the time t, meaning that this is the greatest height of the projectile.

Now we can find the horizontal range of the projectile by using the same equation, but this time in the x-direction.

We will use the x-component of the initial velocity and the acceleration in the x-direction, which is always 0 m/s² assuming air resistance is negligible. Note that the time we have above (25.51 sec) is only half the total time, so we will use double this time (51.02 sec).

The horizontal range of the projectile is 22092.31 m.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

there are no contact forces mentioned on the list of choices that you've provided.