Asked by cassie

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.

Answers

Answered by Tony
Horizontally:
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.
Answered by cassie
thank you so much!
Answered by Anonymous
A particle travels between two parallel vertical
walls separated by 25 m. It moves toward
the opposing wall at a constant rate of
8.8 m/s. It hits the opposite wall at the same
height.

Answered by Chance
If a particle reaches its max height in 15s, what is its range if it is launched at a speed of 275ms that remains constant throughout its flight at angle of 0?
Anwser:
8250m
Answered by Chance
Hi i'm in 4th.
Answered by Chance
It's just I do something like this sometimes.
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