Question

A stone is thrown at an angle of 30° above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A stop watch measures the stone's trajectory time from top of cliff to bottom to be 5.6 s. How far out from the cliff's edge does the stone travel horizontally?

Answer 1

The stone, thrown from a cliff with an initial speed of 12 m/s at an angle of 30°, travels 58.2 meters horizontally from the cliff's edge after 5.6 seconds in flight, calculated using projectile motion principles.

Step-by-step explanation:

The distance a stone travels horizontally after being thrown from a cliff can be determined using principles of projectile motion. Given an initial speed of 12 m/s at an angle of 30° above the horizontal and a flight time of 5.6 seconds, we can calculate the horizontal distance using the following steps:

• Resolve the initial velocity into horizontal and vertical components.
• Use the horizontal component (vx) for distance calculation as it remains constant in the absence of air resistance.
• The horizontal distance traveled is then vx multiplied by the time of flight.

Firstly, we calculate the horizontal velocity component (vx) with vx = v * cos(θ), where v is the initial velocity and θ is the launch angle. Plugging in the numbers, vx = 12 * cos(30°) = 12 * (√3/2) = 10.392 m/s. Then, the horizontal distance (d) is vx * t, where t is the time. So, d = 10.392 m/s * 5.6 s = 58.1952 m.

Therefore, the stone travels 58.2 meters horizontally from the cliff's edge.