What will be its speed when it hits the opposing wall?

Question

A particle travels horizontally between two parallel walls separated by 18.4m. It moves toward the opposing wall at a constant rate of 7.7 m/s. It also has an acceleration in the direction parallel to the walls of 2.9 m/s2.

a. what will be its speed when it hits the opposing wall.

b. At what angle with the wall will the particle strike?

Answer 1

Let us say that the particle touches one wall at t = 0, and bounce out to the other wall. Initially it does not have velocity parallel to the wall. It will hit the other wall then at:
t = 18.4/7.7 (s)
thus its velocity parallel to the wall at t = 18.4/7.7 s is: at = 2.9*18.4/7.7 (m/s).
(a) when the particle hits the opposing wall, its speed is:
S = sqrt {7.7^2 + (2.9*18.4/7.7)^2}
(b) the angle can be calculated with:
sin(the angle) = at /S
the angle = sin^(-1) {(2.9*18.4/7.7 )/ sqrt [7.7^2 + (2.9*18.4/7.7)^2]}
Go ahead with numerical calculation, please.