A particle travels between two parallel vertical walls separated by 11 m. It moves toward the opposing wall at a constant rate of 5.2 m/s. It hits the opposite wall at the same height.
The acceleration of gravity is 9.8 m/s2.
a) What will be its speed when it hits the
opposing wall?
Answer in units of m/s
b)At what angle with the wall will the particle strike?
To find the speed when the particle hits the opposing wall, we can use the concept of projectile motion. Here's how you can calculate it:
a) The vertical distance traveled by the particle can be calculated using the equation:
\[ d = \frac{1}{2} \times g \times t^2 \]
Given that the particle travels a total horizontal distance of 11 m and the acceleration due to gravity is 9.8 m/s^2, we can use the vertical distance as the height between the two walls.
The time taken, \( t \), can be calculated using the equation:
\[ t = \frac{d}{v} \]
where \( v \) is the vertical velocity of the particle.
Since the particle's speed is constant and there is no vertical acceleration, the time taken for the particle to travel across 11 m horizontally is the same as the time taken for it to fall a vertical distance equal to the height of the walls.
So, substituting the values into the equations, we have:
\[ t = \frac{11}{5.2} \]
Now, we can find the vertical velocity, \( v \), using the equation:
\[ v = g \times t \]
Substituting the values, we get:
\[ v = 9.8 \times \frac{11}{5.2} \]
Therefore, the speed of the particle when it hits the opposing wall is \( v \). You can substitute the values in the equation to find the exact value.
b) To find the angle at which the particle strikes the wall, we can use trigonometry. Since the particle hits the wall at the same height, the vertical component of its velocity will be the same as the initial vertical velocity.
Using this vertical component, we can calculate the angle using the equation:
\[ \theta = \tan^{-1} \left(\frac{v}{5.2}\right) \]
Substituting the value of \( v \) obtained from part (a) into the equation will give you the angle at which the particle strikes the wall.
a) To find the speed when the particle hits the opposing wall, we can use the equation:
Distance = Speed x Time
The particle travels a distance equal to the separation between the walls, which is 11 m. The speed is given as 5.2 m/s. We need to find the time it takes for the particle to reach the opposite wall.
Time = Distance / Speed
Time = 11 m / 5.2 m/s
Time ≈ 2.115 seconds
Now, we can find the particle's final speed using the equation for free fall motion:
Final Speed = Initial Speed + (Acceleration x Time)
The initial speed is 0 m/s since the particle starts from rest. The acceleration is the acceleration due to gravity, which is -9.8 m/s^2 (negative because the direction is downward).
Final Speed = 0 m/s + (-9.8 m/s^2) x 2.115 s
Final Speed ≈ -20.722 m/s
However, we are only interested in the magnitude of the velocity, so we take the absolute value:
Final Speed ≈ 20.722 m/s
Therefore, the speed when the particle hits the opposing wall is approximately 20.722 m/s.
b) To find the angle at which the particle strikes the wall, we can use trigonometry. The angle can be determined by finding the inverse tangent of the vertical velocity component divided by the horizontal velocity component.
Vertical Velocity Component = Final Speed x sin(θ)
Horizontal Velocity Component = Final Speed x cos(θ)
Since the vertical velocity component is in the opposite direction of gravity, we consider it as negative. The horizontal velocity component remains positive.
Vertical Velocity Component = -20.722 m/s x sin(θ)
Horizontal Velocity Component = 20.722 m/s x cos(θ)
Using these equations, we can find the angle θ:
tan(θ) = Vertical Velocity Component / Horizontal Velocity Component
tan(θ) = (-20.722 m/s x sin(θ)) / (20.722 m/s x cos(θ))
tan(θ) = -sin(θ) / cos(θ)
Next, we solve for θ:
θ = arctan(-sin(θ) / cos(θ))
This equation gives us the angle at which the particle strikes the wall. Please note that the result will depend on the initial conditions, specifically the angle at which the particle was launched.