A jai alai court is typically 176 ft long between vertical end walls 40 ft high.

Question

A jai alai court is typically 176 ft long between vertical end walls 40 ft high. In the game, balls are thrown against a front wall with a hand-held “cesta”. The balls used are so made that they collide with the walls elastically, just like billiard balls off the edges of a snooker table. A shot (called a “chula shot”), a player would like to make because of the difficulty for the opposing player in judging its rebound, is off the front wall and back to where the floor and back wall meet, as shown in the figure (not to scale). If the shot also reaches the maximum height, it is less likely to be intercepted before reaching the rear wall; this height is limited by a horizontal ceiling screen which is mounted 50 ft above the floor. If such a shot is made starting practically at floor level (not realistic but it makes the problem easier) and at a point 52 ft in front of the rear wall, at what velocity does it leave the cesta?

*Note that the motion of the ball should be split into two projectile motions, one before the collision with the front wall and the other after.

Explanation / Answer

Total Horizontal Distance travlled by the ball = 124 ft + 176 ft = 300 ft

Let the time taken for the whole motion of the Ball be T.

At T/2, the Ball must have attained the Maximum Height.

Also, let the Velocity Component in the Horizontal Direction be Vx

Vx remains constant throughout, as there is no force acting in the Horizonal direction.

Vx* T = 300ft

Also, Maximum Height attained, H = 50 ft

Time Taken to attain Max. height, = T/2 s

Let the Initial Velocity in the Vertical Direction be Vy(0)

Velocity in the vertical direction at Maximum Height is 0.

H = Vy(0)* T/2 + 1/2 * g * (T/2)2, where g = -32.174 ft/s2

Also, using v = u +a*t,

we obtain, 0 = Vy(0) + g*T/2

-> Vy(0) = - g * T/2

-> H =  - g * T/2 * T/2 + 1/2 * g * (T/2)2, where g = -32.174 ft/s2

-> H = - 1/2 g * T2/4

-> 50 ft = -1/8 * - 32.174* T2

T2 = 400/32.174 s2 = 12.432 s2

T = 3.526 s

Vy(0) = - g * T/2 = 32.174 * 3.526/2 = 52.7227 ft/s

Vx* T = 300ft

Vx = 300ft/3.526s = 85.082 ft/s

Let the Vertical Velocity at the time of Collsion be Vc.

Using the relation of v2- u2= 2as,

Vc2 = Vy(0)2 + 2* g * 40ft

= 52.72272 + 2* (-32.174) * 40

= 205.763

Vc = 14.34 ft/s

The Velocity with which the ball leaves the cesta = (Vc2+ VX2) = 86.281 ft/s

Hence, the Required Velocity is 86.281 ft/s

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