9.

A stone is thrown at an angle of 30° above the horizontal from the top edge of a cliff with an initial speed of 19.6 m/s. A stop watch measures the stone's trajectory time from top of cliff to bottom to be 6.7 s. How far out from the cliff's edge does the stone travel horizontally? (g = 9.8 m/s2 and air resistance is negligible)
A) 58 m
B) 114 m
C) 120 m
D) 175 m
E) 197 m

Multiply the horizontal velocity component (which remains constant) by the trajectory time and you will have the horizontal distance traveled. Round to the nearest meter and it will agree with one of the choices.

154.3

To find the horizontal distance traveled by the stone, we need to determine the time it takes for the stone to reach the bottom of the cliff and the initial vertical velocity.

Step 1: Decompose the initial velocity
The initial velocity of the stone can be decomposed into its horizontal and vertical components.

The horizontal component of the initial velocity is given by:
Vx = V0 * cos(theta)
where V0 is the initial speed and theta is the angle of projection.

The vertical component of the initial velocity is given by:
Vy = V0 * sin(theta)

Step 2: Find the time of flight
The time of flight (total time the stone is in the air) can be determined using the vertical component of velocity. The equation for the vertical motion is:
Y = V0y*t - (1/2) * g * t^2

Since the stone starts and ends at the same vertical position, Y = 0. Therefore, the equation simplifies to:
0 = V0y*t - (1/2) * g * t^2

Rearranging the equation, we have:
(1/2) * g * t^2 = V0y*t

Solving for t, we get:
t = 2 * V0y / g

Step 3: Find the horizontal distance
The horizontal distance (range) is given by:
R = Vx * t

Substituting the values and solving, we get:
R = (V0 * cos(theta)) * (2 * V0 * sin(theta) / g)

Simplifying further, we have:
R = (2 * V0^2 * sin(theta) * cos(theta)) / g

Plugging in the given values:
V0 = 19.6 m/s
theta = 30°
g = 9.8 m/s^2

R = (2 * 19.6^2 * sin(30°) * cos(30°)) / 9.8

Using the trigonometric identity sin(2θ) = 2sin(θ)cos(θ), we can simplify further:
R = (19.6^2 * sin(60°)) / 9.8

Since sin(60°) = √3/2, we have:
R = (19.6^2 * (√3/2)) / 9.8

Solving for R, we get:
R ≈ 114 m

Therefore, the stone travels approximately 114 m horizontally from the cliff's edge.

So the correct answer is option B) 114 m.

To solve this problem, we can break down the stone's motion into horizontal and vertical components:

1. Vertical Motion:
We know that the stone is thrown at an angle of 30° above the horizontal. The initial vertical velocity (Vy) can be calculated using the formula: Vy = V0 * sin(theta), where V0 is the initial speed and theta is the angle of projection. Plugging in the values, we get:
Vy = 19.6 m/s * sin(30°)
Vy = 9.8 m/s

Next, we can calculate the time it takes for the stone to reach the ground using the vertical component of motion. The formula to calculate time (t) is: t = 2 * Vy / g, where g is the acceleration due to gravity. Substituting the values, we get:
t = 2 * 9.8 m/s / 9.8 m/s^2
t = 2 s

2. Horizontal Motion:
The horizontal distance traveled by the stone can be calculated using the formula: distance = V0 * cos(theta) * time, where V0 is the initial speed, theta is the angle of projection, and time is the total time of flight. Plugging in the values, we get:
distance = 19.6 m/s * cos(30°) * 6.7 s
distance = 19.6 m/s * sqrt(3) / 2 * 6.7 s
distance = 197 m

Therefore, the stone travels horizontally 197 meters from the cliff's edge.

The correct answer is E) 197 m.