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.
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.
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
Hi i'm in 4th.
It's just I do something like this sometimes.
To find the speed when the particle hits the opposing wall, we can use the equations of motion. Since the particle has a constant acceleration, we can use the equation:
v^2 = u^2 + 2as
where:
v is the final velocity (which we have to find),
u is the initial velocity (5.5 m/s),
a is the acceleration (1.8 m/s^2), and
s is the displacement between the two walls (18.4 m).
Plugging in the values, we can solve for v:
v^2 = (5.5)^2 + 2(1.8)(18.4)
v^2 = 30.25 + 66.24
v^2 = 96.49
v ≈ 9.82 m/s
Therefore, the speed of the particle when it hits the opposing wall is approximately 9.82 m/s.
Now, to find the angle at which the particle will strike the wall, we can use trigonometry. Let's assume the angle is θ.
We can use the equation:
tan(θ) = perpendicular distance / horizontal distance
The perpendicular distance is the height of the wall, which is typically not given in the problem statement. To solve for θ, we need to determine the perpendicular distance.
If we consider the displacement s between the walls as the hypotenuse of a right triangle, we can find the horizontal distance and the perpendicular distance using trigonometry. The horizontal distance is simply the width of the walls (18.4 m), and the perpendicular distance is the height of the wall.
tan(θ) = perpendicular distance / 18.4
Rearranging the equation, we get:
perpendicular distance = tan(θ) * 18.4
Since the height of the wall is not provided in the problem statement, we cannot find the exact angle at which the particle will strike. However, if we know the height of the wall or assume it to be a certain value, we can substitute that value into the equation to find the angle θ.
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.