The constant 7.2 m/s velocity is the component perpendicular to the walls. Call it Vx. The particle only accelerates in the direction parallel to the walls.
The time it takes to cross from one wall t the other is
D/Vx = (18.4 m)/(7.2 m/s) = 2.56 s
In that time, it will have acquired a vertical velocity component of
Vy = (1/2)*2*(2.56)^2 = 6.53 m/s
The speed when it hits the wall is
V = sqrt[(6.53)^2 + (7.2)^2] = 9.72 m/s
The velocity has an an angle at impact, with respect to the horizontal, of
arctan (Vy/Vx) = arctan(.907) = 42.2 degrees
A particle travels horizontally between two parallel walls separated by 18.4m. It moves towards the opposing wall at a constant rate of 7.2m/s. Also, it has an acceleration in the direction parallel to the walls of 2 m/s^2.
a) What will its speed be when it hits the wall?
b) At what angle with the wall will the particle strike?
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