posted by cassie on .
A particle travels horizontally between 2 parallel walls separated by 18.4 m. It moves toward the opposing wall at a constant rate of 5.5 m/s. Also it has an acceleration in the direction parallel to the walls of 1.8 m/s^2.What will its speed be when it hits the opposing wall? answer in m/s.
At what angle with the wall will the particle strike? Answer in units of degrees.
Vavg = x/t so t x/Vavg
t = 18.4 m/5.5m/s = 3.345 sec
Vertically: (parallel to the wall)
Assuming initial velocity in this direction is zero
Vfinal = Vinitial + at so:
Vfinal = 0 + (1.8 m/s^2)(3.345 sec) = 6.0 m/s
Using pythagorean theorem to find the final velocity and tangent to find the angle:
Square root (5.5 ^2 + 6^2) = 8.14 m/s
tan x = 6/5.5 = 47.5 degree with respect to the horizontal so 42.5 degrees with respect to the wall.
thank you so much!