please show me how to do this in steps.
A track star in the long jump goes into the jump at 12 m/s and launches herself at 20.0degrees above the horizontal. What is the magnitude of her horizontal displacement?
(assume no air resistance and that
Ay= -g = -9.81 m/s^2)
thanks :)
Her vertical component of velocity at takeoff is
Vyo = 12 sin 20 = 4.10 m/s
and her horizontal component is
Vx = 12 cos 20 = 11.28 m/s.
The horizontal component remains constant during the jump.
Use the vertical velocity component to determine how long she is in the air before hitting the ground. Call that time T
Vo = g T/2
T = 2 Vo^2/g
The length of the jump is Vx*T
= 2 (Vo^2/g) sin 20 cos 20
= (Vo^2/g) sin 40
Sure! Here are the steps to find the magnitude of the horizontal displacement:
Step 1: Identify the given information:
- Initial velocity in the x-direction (horizontal) = 12 m/s
- Launch angle = 20.0 degrees
- Acceleration due to gravity (Ay) = -9.81 m/s²
Step 2: Determine the initial velocity components:
- Velocity in the x-direction (Vx) = V * cos(theta)
Vx = 12 m/s * cos(20.0 degrees)
Vx = 12 m/s * 0.9397
Vx = 11.2764 m/s
- Velocity in the y-direction (Vy) = V * sin(theta)
Vy = 12 m/s * sin(20.0 degrees)
Vy = 12 m/s * 0.3420
Vy = 4.1040 m/s
Step 3: Find the time of flight (t):
- The time it takes for the object to reach the ground is given by the equation:
t = 2 * Vy / Ay (from the vertical motion equations)
t = 2 * 4.1040 m/s / -9.81 m/s²
t = -0.8377 s (Ignore negative value since time cannot be negative in this context)
Step 4: Calculate the horizontal displacement (x):
- The horizontal displacement can be found using the equation:
x = Vx * t
x = 11.2764 m/s * -0.8377 s
x = -9.4331 m
Step 5: Determine the magnitude of the horizontal displacement:
- The magnitude is the absolute value of the displacement, so:
Magnitude of horizontal displacement = |x|
Magnitude of horizontal displacement = |-9.4331 m|
Magnitude of horizontal displacement = 9.4331 m
Therefore, the magnitude of the horizontal displacement of the track star is 9.4331 meters.
To find the magnitude of the track star's horizontal displacement, we can break down this problem into three steps:
Step 1: Find the vertical component of the initial velocity.
Step 2: Find the time of flight.
Step 3: Calculate the horizontal displacement.
Let's go through each step in detail:
Step 1: Find the vertical component of the initial velocity.
The vertical component of the initial velocity can be calculated using trigonometry. The formula is:
Viy = Vi * sin(θ)
where Viy is the vertical component of the initial velocity, Vi is the initial velocity (12 m/s), and θ is the launch angle (20.0 degrees).
Using the formula:
Viy = 12 m/s * sin(20.0 degrees)
Viy ≈ 4.10 m/s
Step 2: Find the time of flight.
The time of flight is the total time the track star is in the air. We can find it using the vertical component of the initial velocity.
The formula for the time of flight is:
T = (2 * Viy) / g
where T is the time of flight, Viy is the vertical component of the initial velocity (4.10 m/s), and g is the acceleration due to gravity (-9.81 m/s^2).
Using the formula:
T = (2 * 4.10 m/s) / -9.81 m/s^2
T ≈ 0.836 s
Step 3: Calculate the horizontal displacement.
The horizontal displacement can be determined using the horizontal component of the initial velocity and the time of flight.
The formula for horizontal displacement is:
Dx = Vix * T
where Dx is the horizontal displacement, Vix is the horizontal component of the initial velocity, and T is the time of flight.
To find Vix, we use the formula:
Vix = Vi * cos(θ)
where Vix is the horizontal component of the initial velocity, Vi is the initial velocity (12 m/s), and θ is the launch angle (20.0 degrees).
Using the formula:
Vix = 12 m/s * cos(20.0 degrees)
Vix ≈ 11.09 m/s
Now, using the formula for horizontal displacement:
Dx = (11.09 m/s) * (0.836 s)
Dx ≈ 9.26 m
So, the magnitude of the track star's horizontal displacement is approximately 9.26 meters.