A golf ball is given an initial speed of 20 m/s and returns to ground level. Which launch angle above level ground results in the ball traveling the greatest horizontal distance? Neglect friction.
1. 60 degrees
2. 45
3. 30
4. 15
The answer is 45 but I was wondering how to get there. The process.
Thanks.
Well, let me tee up the explanation for you. When considering the launch angle, we need to assess the projectile's motion in two dimensions - the vertical and the horizontal.
Let's slice through the air with some analysis. When the ball is launched at an angle of 45 degrees, it splits the initial velocity between the vertical and horizontal directions equally. So, 20 m/s is evenly distributed as 10 m/s in each direction.
Now, let's focus on the horizontal component while keeping our feet on the ground. Since there is no friction, the horizontal velocity remains constant throughout the ball's flight.
Here comes the punchline - when the golf ball reaches the ground, it takes the same amount of time to come back down as it took to go up. Since gravity pulls it downward at the same rate it was launched upward, the time it spends descending equals the time it spent ascending.
And since we're assuming no air resistance, the horizontal distance traveled by the ball is directly proportional to the time it spends in the air. Therefore, the greatest horizontal distance occurs when the ball spends the same amount of time going up as it does going down.
At an angle of 45 degrees, the time spent going up and coming down is equal. As a result, the horizontal distance covered by the ball will be maximized.
So, the answer to your question is angle number 2, 45 degrees. It's the perfect combination of power and aim, maximizing the horizontal distance the ball travels.
To determine the launch angle that results in the golf ball traveling the greatest horizontal distance, you need to understand the concept of projectile motion.
Projectile motion is the motion of an object that is launched into the air and is subject only to the force of gravity and air resistance (which we'll assume to be negligible in this case).
In projectile motion, the horizontal and vertical components of motion are independent of each other.
To maximize the horizontal distance traveled by the golf ball, you need to maximize the horizontal component of its initial velocity.
The horizontal component of the initial velocity (Vx) remains constant throughout the motion, as there are no horizontal forces acting on the ball.
The horizontal distance traveled by the golf ball is given by the formula: distance = Vx * time.
To find the time of flight (T), you can use the formula for vertical motion:
T = (2 * Vy0) / g
Where Vy0 is the initial vertical component of the velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2).
The initial vertical component of the velocity (Vy0) can be calculated using the formula:
Vy0 = V * sin(θ)
Where V is the initial velocity (20 m/s) and θ is the launch angle.
Now, substitute the value of Vy0 in the equation for T:
T = (2 * V * sin(θ)) / g
The total horizontal distance traveled by the golf ball can be calculated using the formula:
distance = Vx * T
Since Vx is constant and T is the same for all launch angles, the horizontal distance is directly proportional to the initial horizontal velocity component Vx.
To maximize the horizontal distance, you need to maximize Vx.
Since Vx = V * cos(θ), where V is the initial velocity and θ is the launch angle, it becomes clear that the horizontal component of the initial velocity is maximized when the launch angle is 45 degrees.
To determine the launch angle that results in the golf ball traveling the greatest horizontal distance, we need to analyze the projectile motion.
Projectile motion consists of two independent motions: horizontal motion (along the x-axis) and vertical motion (along the y-axis). Neglecting friction, only the vertical component of velocity changes due to gravity, while the horizontal component remains constant.
The horizontal distance traveled by the golf ball is determined by the time of flight and the horizontal velocity. The equation for the horizontal distance is given by:
Range = (initial horizontal velocity) * (time of flight)
In this case, the initial horizontal velocity of the ball will be determined by the initial speed and the launch angle. The initial horizontal velocity can be found using the trigonometric function cosine:
Initial horizontal velocity = (initial speed) * cos(θ)
The time of flight can be calculated using the equation:
time of flight = (2 * initial vertical velocity) / (acceleration due to gravity)
In this case, the initial vertical velocity can be found using the trigonometric function sine:
Initial vertical velocity = (initial speed) * sin(θ)
Now, let's analyze each option mentioned:
1. 60 degrees:
Initial horizontal velocity = (20 m/s) * cos(60) = 10 m/s
Initial vertical velocity = (20 m/s) * sin(60) = 17.32 m/s
The time of flight can be calculated using the equation: time of flight = (2 * 17.32 m/s) / (9.8 m/s^2) ≈ 3.53 seconds
The horizontal distance traveled will be: Range = (10 m/s) * (3.53 seconds) ≈ 35.3 meters
2. 45 degrees:
Initial horizontal velocity = (20 m/s) * cos(45) = 14.14 m/s
Initial vertical velocity = (20 m/s) * sin(45) = 14.14 m/s
The time of flight can be calculated using the equation: time of flight = (2 * 14.14 m/s) / (9.8 m/s^2) ≈ 2.88 seconds
The horizontal distance traveled will be: Range = (14.14 m/s) * (2.88 seconds) ≈ 40.8 meters
3. 30 degrees:
Initial horizontal velocity = (20 m/s) * cos(30) = 17.32 m/s
Initial vertical velocity = (20 m/s) * sin(30) = 10 m/s
The time of flight can be calculated using the equation: time of flight = (2 * 10 m/s) / (9.8 m/s^2) ≈ 2.04 seconds
The horizontal distance traveled will be: Range = (17.32 m/s) * (2.04 seconds) ≈ 35.3 meters
4. 15 degrees:
Initial horizontal velocity = (20 m/s) * cos(15) ≈ 19.33 m/s
Initial vertical velocity = (20 m/s) * sin(15) ≈ 5 m/s
The time of flight can be calculated using the equation: time of flight = (2 * 5 m/s) / (9.8 m/s^2) ≈ 1.02 seconds
The horizontal distance traveled will be: Range = (19.33 m/s) * (1.02 seconds) ≈ 19.7 meters
From the calculations, it is evident that the launch angle of 45 degrees results in the greatest horizontal distance, approximately 40.8 meters. Therefore, the correct answer is option 2.
If one ignores air resistance, then
horizontal distance= V*cosTheta * t
but t is dependent on time in the air.
vertical distance= VsinTheta*t-1/2 g t^2=0
or t= 2VsinTheta/g
putting that into horizontal distance
distance=2V^2cosThetasinTheta/g
but cosThetasinTheta=1/2 sin(2theta)
d=2V/g *sin(2theta)
to find max distance, we take the first derivative and set to zero
d/dtheta (2V/g*cos(2theta)*2)=0
or 2theta=90 degrees, or theta=45 deg for max. For min distance, theta=0