A projectile is projected with speed u at an angle alpha on an inclined plane of inclination angle bita. Find time of flight nn inclined plane and maximum height attained by projectile w.r.t. Inclined plane, range of projectile on inclined plane, angle of projection for which range is maximum and two angle at which range will be same?

To find the time of flight on an inclined plane, maximum height attained, range, angle of projection for maximum range, and two angles at which the range will be the same, follow the steps below:

Step 1: Resolve the initial velocity u into horizontal and vertical components.
The horizontal component of velocity, Vx = u * cos(alpha)
The vertical component of velocity, Vy = u * sin(alpha)

Step 2: Resolve the acceleration due to gravity (g) into the components parallel and perpendicular to the inclined plane.
The component of gravity parallel to the inclined plane, g_parallel = g * sin(beta)
The component of gravity perpendicular to the inclined plane, g_perpendicular = g * cos(beta)

Step 3: Calculate the time of flight on the inclined plane (T).

Using the equation of motion:
Vy = u * sin(alpha) - g_parallel * T
0 = u * sin(alpha) - g_parallel * T

Solving for T:
T = u * sin(alpha) / g_parallel

Step 4: Calculate the maximum height attained by the projectile w.r.t. the inclined plane (H).

Using the equation of motion:
Vy^2 = (u * sin(alpha))^2 - 2 * g_parallel * H
0 = (u * sin(alpha))^2 - 2 * g_parallel * H

Solving for H:
H = (u^2 * sin^2(alpha)) / (2 * g_parallel)

Step 5: Calculate the range of the projectile on the inclined plane (R).

Using the equation of motion:
Vx = u * cos(alpha) - g_perpendicular * T
R = Vx * T = (u * cos(alpha) - g_perpendicular * T) * T
R = u * cos(alpha) * T - g_perpendicular * T^2

Step 6: Calculate the angle of projection for which the range is maximum.

Differentiate R with respect to alpha and set it equal to zero to find the angle of projection for maximum range.

dR/d(alpha) = -u * sin(alpha) * T + g_perpendicular * T^2 = 0

Solving for alpha:
u * sin(alpha) * T = g_perpendicular * T^2
u * sin(alpha) = g_perpendicular * T
sin(alpha) = (g_perpendicular * T) / u

Step 7: Calculate the two angles at which the range will be the same.

Since the range will be the same for mirror angles, let's consider alpha and (180 - alpha). After substituting these values in the equation for R, both expressions will be equal.

R(alpha) = u * cos(alpha) * T - g_perpendicular * T^2
R(180 - alpha) = u * cos(180 - alpha) * T - g_perpendicular * T^2

Simplify both expressions and set them equal to each other:
cos(alpha) = (-cos(alpha))
alpha = 45 degrees

Therefore, the two angles at which the range will be the same are 45 degrees and (180 - 45) = 135 degrees.

To find the time of flight on an inclined plane, maximum height attained by the projectile with respect to the inclined plane, range of the projectile on the inclined plane, the angle of projection for which the range is maximum, and the two angles at which the range will be the same, we need to use the equations of motion and some basic trigonometry.

1. Time of Flight on Inclined Plane:
The time of flight (T) on an inclined plane can be calculated using the formula:

T = (2 * u * sin(alpha)) / g

where:
u is the initial speed of the projectile,
alpha is the angle of projection, and
g is the acceleration due to gravity.

2. Maximum Height Attained by Projectile:
The maximum height attained by the projectile (H) with respect to the inclined plane can be calculated using the formula:

H = (u^2 * sin^2(alpha)) / (2 * g * (cos^2(bita)))

where:
u is the initial speed of the projectile,
alpha is the angle of projection, and
bita is the angle of inclination of the inclined plane.

3. Range of Projectile on Inclined Plane:
The range (R) of the projectile on the inclined plane can be calculated using the formula:

R = (u^2 * sin(2 * alpha)) / g

where:
u is the initial speed of the projectile,
alpha is the angle of projection, and
g is the acceleration due to gravity.

4. Angle of Projection for Maximum Range:
To find the angle of projection (alpha) for which the range is maximum, differentiate the range equation with respect to alpha and equate it to zero:

dR/d(alpha) = (2 * u^2 * cos(alpha) * sin(alpha)) / g = 0

Simplifying the equation, we get:
cos(alpha) * sin(alpha) = 0

This equation implies that either cos(alpha) or sin(alpha) should be zero. Therefore, the angle of projection for which the range is maximum is when alpha = 0 degrees or alpha = 90 degrees.

5. Two Angles at Which Range will be Same:
To find the two angles at which the range will be the same, set the expressions for the range with different angles equal to each other:

(u^2 * sin(2 * alpha1)) / g = (u^2 * sin(2 * alpha2)) / g

Canceling out constants, we get:
sin(2 * alpha1) = sin(2 * alpha2)

Now, there are two possibilities here:
- Either the values inside the sine function are equal, i.e., (2 * alpha1) = (2 * alpha2). In this case, the two angles will be the same.
- Or the values inside the sine function are supplementary, i.e., (2 * alpha1) + (2 * alpha2) = 180 degrees. In this case, the two angles will be supplementary (add up to 180 degrees).

By using these equations and formulas, you can calculate the time of flight, maximum height, range, angle of projection for maximum range, and the two angles at which the range will be the same.