For a science fair competition, a group of High School students built a kicker-machine that can launch a golf ball from the origin with a velocity of 12.3 m/s and initial angle of 29.9° with respect to the horizontal.
*Where will the golf ball fall back to the ground?
*How High will it be at it's highest point of trajectory?
*What is the ball's velocity vector (in Cartesian components) at the highest point of its trajectory? x, then y
*What is the ball's acceleration vector (in Cartesian components) at the highest point of its trajectory? x, then y
Answer the last two questions immediately.
At the highest point the ball has no y velocity and the x velocity is the same for the whole trip, u = 12.3 cos 29.9
During the whole trip the acceleration is only in the y direction and is -9.8m/s^2
Now do the problem.
Y(vertical) problem first.
Vi = 12.3 sin 29.9
at top v = 0
0 = Vi - 9.8 t
9.8 t = 12.3 sin 29.9
solve for t at top, which is half the trip time
h = Vi t - 4.9 t^2
solve for h, height at top
T = 2 t, total time in air
then d = u T = [12.3 cos 29.9 ] * T
thanks a lot@Damon
To answer your questions, we will use the equations of motion for projectile motion.
1. Where will the golf ball fall back to the ground?
To find the horizontal range, we can use the formula for horizontal displacement:
Range = (initial velocity * time of flight) * cosine(angle)
The time of flight can be calculated using the formula:
time of flight = (2 * initial velocity * sin(angle)) / acceleration due to gravity
Given:
Initial velocity (u) = 12.3 m/s
Angle (θ) = 29.9°
Acceleration due to gravity (g) = 9.8 m/s^2
First, we need to convert the angle to radians:
θ_radians = θ * (π / 180)
Now we can calculate the time of flight:
time of flight = (2 * 12.3 * sin(θ_radians)) / 9.8
Next, we can calculate the horizontal displacement:
Range = (12.3 * time of flight) * cos(θ_radians)
2. How high will it be at its highest point of trajectory?
The maximum height (h) can be calculated using the formula:
h = (initial velocity * sin(angle))^2 / (2 * acceleration due to gravity)
h = (12.3 * sin(θ_radians))^2 / (2 * 9.8)
3. What is the ball's velocity vector (in Cartesian components) at the highest point of its trajectory? x, then y
The vertical component of velocity remains constant throughout the flight (ignoring air resistance), so the y-component at the highest point will be equal to the y-component of the initial velocity.
Vy = initial velocity * sin(angle)
The horizontal component of velocity at the highest point becomes zero.
Vx = initial velocity * cos(angle)
4. What is the ball's acceleration vector (in Cartesian components) at the highest point of its trajectory? x, then y
At the highest point of its trajectory, the ball experiences only vertical acceleration due to gravity, while the horizontal acceleration remains zero.
Ax = 0
Ay = -acceleration due to gravity
Using the provided values, you can plug them into the equations above to find the respective answers to your questions.
To determine the answers to these questions, we can break down the problem into different components and use equations of motion. Let's go step by step:
1. Where will the golf ball fall back to the ground?
To find the horizontal distance, or range, that the golf ball will travel, we can use the horizontal component of the initial velocity. The horizontal component is given by:
Vx = V * cos(θ)
where Vx is the horizontal component of the velocity, V is the magnitude of the velocity (12.3 m/s in this case), and θ is the angle (29.9°).
Using this equation, we can calculate the horizontal component of the initial velocity:
Vx = 12.3 m/s * cos(29.9°) = 10.88161 m/s
Next, we can use the equation of motion for horizontal motion to find the time it takes for the golf ball to hit the ground:
Range = Vx * t
where Range is the horizontal distance and t is the time of flight.
Since we are interested in the point where the ball hits the ground, we can set the height (y) to zero. The equation for vertical motion is given by:
y = Vy * t - (1/2) * g * t^2
where Vy is the vertical component of the velocity and g is the acceleration due to gravity (approximately 9.8 m/s^2).
At the maximum height of the trajectory, the vertical component of the velocity becomes zero (Vy = 0). Hence, we can solve the equation for time t and substitute it back into the equation for horizontal motion to find the range.
2. How high will it be at its highest point of trajectory?
To find the maximum height of the golf ball's trajectory, we can use the equation for vertical motion. At the highest point, the vertical component of the velocity is zero. So, we have:
Vy = V * sin(θ)
where Vy is the vertical component of the velocity, V is the magnitude of the velocity (12.3 m/s), and θ is the angle (29.9°).
Using this equation, we can calculate the vertical component of the initial velocity:
Vy = 12.3 m/s * sin(29.9°) = 6.09658 m/s
Next, we can find the time taken to reach the highest point of the trajectory using the equation:
Vy = Vy_initial - g * t
where Vy_initial is the initial vertical component of the velocity (6.09658 m/s) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
By rearranging this equation, we can solve for the time t it takes to reach the maximum height.
Finally, we can substitute the value of t back into the equation for vertical motion to find the maximum height.
3. What is the ball's velocity vector (in Cartesian components) at the highest point of its trajectory (x, then y)?
The velocity vector at the highest point of the trajectory will have its x-component unchanged since there is no horizontal acceleration. So, we can use the initial horizontal component of the velocity:
Vx = 10.88161 m/s (calculated earlier)
The y-component of the velocity at the highest point will be zero since the ball momentarily stops at the highest point. So, the velocity vector at the highest point will be:
Vx = 10.88161 m/s (x-component)
Vy = 0 m/s (y-component)
4. What is the ball's acceleration vector (in Cartesian components) at the highest point of its trajectory (x, then y)?
At the highest point of the trajectory, the only acceleration acting on the ball is the acceleration due to gravity. The acceleration due to gravity always acts vertically downward, so the x-component of the acceleration will be zero. The y-component of the acceleration is equal to the acceleration due to gravity, which is approximately -9.8 m/s^2.
Therefore, the acceleration vector at the highest point of the trajectory will be:
Ax = 0 m/s^2 (x-component)
Ay = -9.8 m/s^2 (y-component)
By following these steps, you should be able to find the answers to the given questions.