How Does Projectile Angle Affect Flight Time on an Inclined Plane?

[cscott]

A projectile is fired up an inclined plane with an initial speed vi and at an angle beta with respect to the incline, which, in
turn, rises at an angle alpha with respect to the horizontal.

(b) Determine the time of flight, tflight from when the projectile is launched to when it strikes the inclined plane.
Express your answer in terms of alpha, beta, vi and g.

(c) Determine a second expression for the time of flight based on the condition that the projectile strikes the inclined
plane with a final velocity that is perpendicular to the plane. Again, express your answer in terms of alpha, beta, vi and
g.

(d) Using your results from parts (b) and (c), find the angle beta and the values for the initial speed, vi , for which the
final velocity is perpendicular to the inclined plane. If your answer surprises you a clever use of dimensional
analysis may provide some insight.

I have my two expressions for time for (b) and (c) but I'm not sure how to use them to get the angles or values for initial velocity.

(b)$$t = \frac{2v_i\sin \beta}{g \cos \alpha}$$
(c)$$t = \frac{v_i cos \beta}{g \sin \alpha}$$

 

[semc]

hmm...i have yet to try the question but looks interesting...IF we were to assume angle alpha-beta to be zero which means the projectile is on a flat ground,does tt mean tt when the projectile hits the ground,Vy=0?

 

[kyle8921]

+ \frac{v_i \sin \beta}{g \cos \alpha}

I would first analyze the problem and break it down into its basic components. In this case, we have a projectile being launched at an angle beta with respect to the incline, which rises at an angle alpha with respect to the horizontal. We also know that the initial speed, vi, and the acceleration due to gravity, g, will play a role in the motion of the projectile.

To find the time of flight, tflight, we can use the basic equation of motion for a projectile:

y = y_0 + v_{0y}t - \frac{1}{2}gt^2

Where y is the vertical displacement, y_0 is the initial height (which we can assume to be 0 in this case), v_{0y} is the initial vertical velocity (vi sin beta), and t is time.

We can also use the fact that the projectile will reach its maximum height at the halfway point of its flight, and that the time to reach maximum height is equal to half of the total time of flight. Therefore, we can set the equation for maximum height equal to zero and solve for t/2:

0 = vi\sin \beta (t/2) - \frac{1}{2}gt^2/4

Simplifying and solving for t, we get:

t = \frac{2v_i\sin \beta}{g}

This is the same expression as in part (b) of the question, which gives us the time of flight from launch to impact.

For part (c), we are looking for the condition where the final velocity of the projectile is perpendicular to the inclined plane. This means that the horizontal component of the final velocity will be zero. Using the same basic equation of motion, but now setting y = 0 (since the projectile will hit the inclined plane at ground level), we can solve for t:

0 = vi\cos \beta t - \frac{1}{2}gt^2

Solving for t, we get:

t = \frac{2v_i\cos \beta}{g}

This is the same expression as in part (c) of the question, which gives us the time of flight from launch to when the projectile hits the inclined plane with a final velocity perpendicular to the plane.

To find the angle beta and the values for initial