A golf ball is struck from the ground with an initial velocity of 45 m/s at an angle of

25 degrees. It then returns to the ground.
a) What are the x- and y-components of the initial velocity of the golf ball?
b) How long is the golf ball in the air?
c) What is the horizontal displacement of the golf ball?

a) x-component is ... 45 m/s * cos(25º)

... y-component is ... 45 m/s * sin(25º)

b) time up equals time down ... flight time = 2 * {[45 m/s * sin(25º)] / g}

c) horizontal displacement = flight time * 45 m/s * cos(25º)

A golf ball was struck with an initial velocity of 17 m/s at an angle of 55°. Find the horizontal velocity component.

To solve this problem, we can break down the initial velocity into its x and y components.

a) The x-component of the initial velocity can be found using the formula:

Vx = V * cos(theta)

Where V is the magnitude of the initial velocity (45 m/s) and theta is the launch angle (25 degrees).

Vx = 45 * cos(25)
Vx ≈ 40.745 m/s

So, the x-component of the initial velocity is approximately 40.745 m/s.

The y-component of the initial velocity can be found using the formula:

Vy = V * sin(theta)

Vy = 45 * sin(25)
Vy ≈ 19.11 m/s

So, the y-component of the initial velocity is approximately 19.11 m/s.

b) The time the golf ball is in the air can be found using the formula for time of flight:

t = (2 * Vy) / g

Where g is the acceleration due to gravity (9.8 m/s²).

t = (2 * 19.11) / 9.8
t ≈ 3.898 seconds

So, the golf ball is in the air for approximately 3.898 seconds.

c) The horizontal displacement of the golf ball can be found using the formula:

d = Vx * t

Where Vx is the x-component of the initial velocity and t is the time of flight.

d = 40.745 * 3.898
d ≈ 158.84 meters

So, the horizontal displacement of the golf ball is approximately 158.84 meters.

To solve this problem, we can use the principles of projectile motion. Let's break down the problem step by step:

a) To find the x- and y-components of the initial velocity, we can use trigonometry. The x-component (Vx) of the initial velocity can be found by multiplying the initial velocity (45 m/s) by the cosine of the launch angle (25 degrees):

Vx = 45 m/s * cos(25 degrees)

To find the y-component (Vy) of the initial velocity, we multiply the initial velocity by the sine of the launch angle:

Vy = 45 m/s * sin(25 degrees)

So, the x-component (Vx) of the initial velocity is approximately:
Vx = 45 m/s * cos(25 degrees) ≈ 41.13 m/s

And the y-component (Vy) of the initial velocity is approximately:
Vy = 45 m/s * sin(25 degrees) ≈ 19.22 m/s

b) To determine how long the golf ball is in the air, we can use the y-component of the initial velocity. Since the golf ball returns to the ground, the time it takes to reach the maximum height will be the same as the time it takes to return to the ground. At the highest point of its trajectory, the y-component of the velocity becomes zero. We can use this information to calculate the time of flight.

The formula for finding time (t) is given by the equation:

Vy = gt - 1/2 * g * t^2

Where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Using this equation, we set Vy to zero and solve for t:

0 = gt - 1/2 * g * t^2

Rearranging the equation, we have:

1/2 * g * t^2 = gt

Simplifying further:

1/2 * t^2 = t

Dividing both sides by t and rearranging, we get:

t = 2 seconds

So, the golf ball is in the air for approximately 2 seconds.

c) Finally, to determine the horizontal displacement of the golf ball, we can use the x-component of the initial velocity and the time of flight. The equation for horizontal displacement (x) is given by:

x = Vx * t

Substituting the values we found:

x = 41.13 m/s * 2 seconds ≈ 82.26 meters

Therefore, the horizontal displacement of the golf ball is approximately 82.26 meters.