Question
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.
At what angle with the wall will the particle strike? Answer in units of degrees.
Answers
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.
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!
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.
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.
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
Anwser:
8250m
Hi i'm in 4th.
It's just I do something like this sometimes.