A golf ball is hit with an initial velocity of 135 feet per second at an angle of 22 degrees above the horizontal. Will the ball clear a 25 foot wide sand trap whose nearest edge is 300 feet from the golfer?

Will be through with all the word problems soon. I have managed to actually work some on my own!

Thanks.

To determine if the ball will clear the sand trap, we need to find the height of the ball when it reaches the nearest edge of the sand trap.

First, we can find the time it takes for the ball to reach the nearest edge of the sand trap. We can use the horizontal component of the initial velocity and the horizontal distance to the sand trap:

Vx = V * cos(angle)
Vx = 135 * cos(22°)
Vx ≈ 126.39 ft/s

Horizontal distance = Vx * t
300 = 126.39 * t
t ≈ 2.37 seconds

Next, we can find the vertical component of the initial velocity:

Vy = V * sin(angle)
Vy = 135 * sin(22°)
Vy ≈ 50.71 ft/s

Now we can find the height of the ball at the nearest edge of the sand trap using the vertical component of the velocity:

y = Vy*t - 0.5*g*t^2
y = 50.71*2.37 - 0.5*32*2.37^2
y ≈ 120.14 feet

Since the height of the ball (120.14 feet) is greater than the 25-foot wide sand trap, the ball will clear the sand trap.

To determine whether the golf ball will clear the sand trap, we can analyze the horizontal and vertical components of the ball's motion.

1. Horizontal motion: The initial velocity of the ball can be split into horizontal and vertical components using trigonometry. The horizontal component, Vx, is given by Vx = V * cos(theta), where V is the initial velocity (135 ft/s) and theta is the launch angle (22 degrees).

Vx = 135 ft/s * cos(22 degrees)
Vx = 126.842 ft/s (rounded to three decimal places)

2. Vertical motion: The vertical component of the ball's velocity can be determined using the equation Vy = V * sin(theta), where Vy is the vertical velocity.

Vy = 135 ft/s * sin(22 degrees)
Vy = 49.095 ft/s (rounded to three decimal places)

3. Time of flight: The time it takes for the golf ball to reach the ground can be determined using the formula t = 2 * Vy / g, where g is the acceleration due to gravity (32.2 ft/s^2).

t = 2 * 49.095 ft/s / 32.2 ft/s^2
t = 3.040 seconds (rounded to three decimal places)

4. Horizontal distance traveled: The horizontal distance the ball travels can be determined using the formula d = Vx * t, where d is the horizontal distance.

d = 126.842 ft/s * 3.040 seconds
d = 385.151 feet (rounded to three decimal places)

Since the horizontal distance traveled by the golf ball is greater than the distance to the nearest edge of the sand trap (300 feet), the ball will clear the sand trap.

To determine whether the golf ball will clear the sand trap, we can use basic kinematic equations to analyze the projectile motion of the golf ball.

First, we need to break down the initial velocity into its horizontal and vertical components. The horizontal component (Vx) represents the velocity in the horizontal direction, and the vertical component (Vy) represents the velocity in the vertical direction.

Given:
Initial velocity (V) = 135 ft/s
Launch angle (θ) = 22 degrees

The horizontal component can be calculated using the formula:

Vx = V * cos(θ)

Vx = 135 ft/s * cos(22 degrees) ≈ 125.26 ft/s

The vertical component can be calculated using the formula:

Vy = V * sin(θ)

Vy = 135 ft/s * sin(22 degrees) ≈ 51.78 ft/s

Next, we can calculate the time it takes for the golf ball to reach the point directly above the sand trap (300 feet away) using the horizontal component velocity:

Time = Distance / Velocity

Time = 300 ft / 125.26 ft/s ≈ 2.395 seconds

Now, we can calculate the maximum height (H) the golf ball will reach using the vertical component velocity:

H = (Vy^2) / (2 * g)

where g is the acceleration due to gravity (32.17 ft/s²).

H = (51.78 ft/s)^2 / (2 * 32.17 ft/s²) ≈ 42.02 ft

Finally, we need to determine if the ball will clear the 25-foot wide sand trap.

To do this, we need to calculate the horizontal distance (D) the golf ball will travel in the air using the horizontal component velocity and the time it takes to reach the point directly above the sand trap:

D = Vx * Time

D = 125.26 ft/s * 2.395 s ≈ 299.76 ft

Since the sand trap's nearest edge is 300 feet away from the golfer, and the ball will travel approximately 299.76 feet in the air, it means the ball will not clear the sand trap.