Where u and v and initial and final velocities

[mcintyre_ie]

Hey
Looking for some help with another projectiles question:

(A)
A particle is projected from the ground with a velocity of 50.96 m/s at an angle Tan-1 5/12 to the horizontal. On its upward path, it just passes over a wall 14.7m high. During its flight it also passes over a second wall, 18.375 m high.
Show that the second wall must not be less than 23.52 m and not more than 70.56 m from the first wall.

(B)
A plane is inclined at an angle of 2B (beta) to the vertical. A particle is projected up the plane with initial velocity uCosB, at an angle B (beta) to the inclined plane. The plane of projection is vertical and contains the line of greatest slope.
Show:
(I) that the time of flight of the particle is u/g
(II)that the range ofthe particle on the plane is u^2/2g

Ok, so for part (A) I've made this sketch:

http://community.webshots.com/s/image1/4/51/1/95445101CnvHBl_ph.jpg

Ux = 47.04
Uy = 19.6
Vx = 47.04
Vy = 19.6 - gt
Ax = 0
Ay = -g
Sx = 47.04t
Sy = 19.6t - 1/2gt^2

where u and v and initial and final velocities, and a and s are acceleration and distance, x and y are in the horizontal and vertical directions.

I subbed in Sy = 14.7 and got a quadratic with two solutions - t=1 and t=3, since the time of flight is 8 seconds, either solution is acceptable.
Im stuck as to what to do next. Any hep would be appreciated.

(B) Heres my sketch for this part:
http://community.webshots.com/s/image1/4/49/1/95444901reiLfQ_ph.jpg

Im totally stuck here. The given velocity is what i would have thought to be the x velocity, suggesting the particle has no y velocity - which can't be possible i think? Any help as to how to break this down would be helpful

 

[mcintyre_ie]

Any help is appreciated

 

[M98Ranger]

.

Hi there, for part (A) you are on the right track. To find the range of the particle, you can use the equation for horizontal distance: Sx = Ux * t. Since the particle passes over the second wall, the horizontal distance traveled must be equal to the distance between the two walls (which we will call d). So, setting Sx = d, we get: d = Ux * t. Substituting in t=1 or t=3 (since both solutions for t are acceptable), we get two equations: d = 47.04 * 1 = 47.04, and d = 47.04 * 3 = 141.12. Therefore, the second wall must be at a distance between 47.04 m and 141.12 m from the first wall. Since the second wall is also 18.375 m high, it must be at a height between 23.52 m and 70.56 m from the first wall.

For part (B), you are correct in thinking that the given velocity is the x component. This means that the particle has no y velocity at the beginning, and will only start to gain a y component as it moves up the inclined plane. To find the time of flight, you can use the equation Sy = Uy * t + 1/2 * Ay * t^2, where Sy is the vertical distance (which we can set to the height of the plane, 18.375 m), Uy is the initial y velocity (which we know is 0), and Ay is the acceleration in the y direction (which we know is -g). Substituting in these values, we get: 18.375 = 0 * t + 1/2 * (-9.8) * t^2. Simplifying, we get t^2 = 3.75, and taking the square root, t = 1.936 seconds. This is the time it takes for the particle to reach the top of the plane. Since the particle will also take the same amount of time to come back down, the total time of flight will be 2 * 1.936 = 3.872 seconds.

To find the range of the particle on the plane, you can use the equation Sx = Ux * t, where Sx is the horizontal distance (which we can set to the