A squash ball typically rebounds from a surface with 25% of the speed with which

Question

A squash ball typically rebounds from a surface with 25% of the speed with which it initially struck the surface. Suppose a squash ball is served in a shallow trajectory, from a height above the floor of 51 cm, at a launch angle of 6.0 degree above the horizontal, and at a distance of 11 m from the front wall. If it strikes the front wall exactly at the top of its parabolic trajectory, determine how above the surface the ball strikes the wall? How far horizontally from the war does it strike the floor, after rebounding? (Ignore any effects due to air resistance.)

Explanation / Answer

a) Here, u * cos6 * t = 11

Also, u * sin6 = gt

=> u * cos6 * t * u * sin6 = 11 * gt

=> u = 30.7 m/sec(approx)

=> high above the surface the ball strikes the wall = 0.51 + 30.72 * sin26/19.6

= 1.035 m

b) far horizontally from the wall does it strike the floor = 0.25 * 30.7 * cos6 * sqrt(2 * 1.035/9.8)

= 3.508 m

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