Suppose that Tiger hit a golf ball with an initial velocity of 150 ft/sec at an angle of elevation of 30°. Write the parametric equations to represent the flight of the ball. How long is the golf ball in the air? Determine the distance the ball traveled. When is the ball at its maximum height? What is the maximum height?

To write the parametric equations to represent the flight of the ball, we can break down the initial velocity into its horizontal and vertical components.

The horizontal component, v_x, is given by v_x = v * cos(theta), where v is the initial velocity and theta is the angle of elevation. In this case, v_x = 150 ft/sec * cos(30°) = 150 ft/sec * sqrt(3)/2 = 75 sqrt(3) ft/sec.

The vertical component, v_y, is given by v_y = v * sin(theta), where v is the initial velocity and theta is the angle of elevation. In this case, v_y = 150 ft/sec * sin(30°) = 150 ft/sec * 1/2 = 75 ft/sec.

Let's denote the horizontal position of the ball as x(t) and the vertical position as y(t) at time t. The parametric equations are then:
x(t) = v_x * t
y(t) = v_y * t - 16t^2, where 16 is half the acceleration due to gravity in ft/sec^2.

To find how long the golf ball is in the air, we need to determine when the ball hits the ground. At the moment it hits the ground, the vertical position y(t) becomes 0. So, we set y(t) = 0 and solve for t.

0 = v_y * t - 16t^2
0 = 75t - 16t^2
0 = t(75 - 16t)
t = 0 (initial time) or t = 75/16

Since time cannot be negative, the ball is in the air for 75/16 seconds.

To determine the distance the ball traveled, we need to find the horizontal distance traveled. This can be calculated by evaluating x(t) at t = 75/16.

x(t) = v_x * (75/16)
x(t) = (75 sqrt(3) ft/sec) * (75/16 sec)
x(t) = (75 * 75 sqrt(3)) / (16) ft
x(t) = 281.25 sqrt(3) ft

So, the ball traveled approximately 281.25 sqrt(3) ft.

The ball reaches its maximum height when its vertical velocity becomes zero. We can find this time by setting v_y - 16t = 0 and solving for t.
v_y - 16t = 0
75 - 16t = 0
16t = 75
t = 75 / 16 seconds

Therefore, the ball reaches its maximum height after approximately 75/16 seconds.

To find the maximum height, we substitute this time into the vertical position equation:

y(t) = v_y * t - 16t^2
y(t) = 75 * (75/16) - 16 * (75/16)^2
y(t) = 281.25 - 281.25/4
y(t) = (4/4) * 281.25 - 281.25/4
y(t) = 3 * 281.25/4
y(t) = 843.75/4 ft
y(t) = 210.94 ft

Therefore, the maximum height of the ball is approximately 210.94 ft.

To write the parametric equations for the flight of the ball, we can use the following equations:

x = v0 * cos(theta) * t
y = v0 * sin(theta) * t - 0.5 * g * t^2

Where:
- v0 is the initial velocity (150 ft/sec)
- theta is the angle of elevation (30°)
- t is the time in seconds
- g is the acceleration due to gravity (32.17 ft/sec^2)

Substituting the given values, the parametric equations become:

x = 150 * cos(30°) * t
y = 150 * sin(30°) * t - 0.5 * 32.17 * t^2

To find the time the ball is in the air, we need to find when y = 0. Setting y equal to 0, we have:

0 = 150 * sin(30°) * t - 0.5 * 32.17 * t^2

Solving this quadratic equation for t, we find two roots:
t = 0 (initial time)
t ≈ 7.94 seconds.

Therefore, the ball is in the air for approximately 7.94 seconds.

To determine the distance the ball traveled, we need to find the value of x when t = 7.94 seconds.

x = 150 * cos(30°) * 7.94
x ≈ 817.89 feet.

Therefore, the ball traveled approximately 817.89 feet.

To find the time when the ball is at its maximum height, we can use the fact that the ball reaches its maximum height when the vertical velocity component becomes zero. This occurs at the peak of the ball's trajectory.

The vertical component of the velocity is given by:

vy = v0 * sin(theta) - g * t

Setting vy equal to zero and solving for t, we have:

0 = 150 * sin(30°) - 32.17 * t

Solving for t, we find:

t ≈ 4.87 seconds.

Therefore, the ball is at its maximum height approximately 4.87 seconds after it was launched.

To determine the maximum height of the ball, we can substitute this time into the y equation:

y = 150 * sin(30°) * 4.87 - 0.5 * 32.17 * (4.87)^2

y ≈ 97.75 feet.

Therefore, the maximum height of the ball is approximately 97.75 feet.

since the horizontal speed does not change, we have the usual

x(t) = 150 cos 30° t
y(t) = 150 sin 30° t - 16 t^2

The rest is downhill from here