A hiker throws a stone from the upper edge of a vertical cliff. The stone's initial velocity is 25.0 m/s directed at 40.0° with the face of the cliff, as shown in the figure.The stone hits the ground 3.75 s after being thrown.How far from the foot of the cliff does the stone land?Ignore air friction and express your answer in meters.

To solve this problem, we need to break it down into horizontal and vertical components.

Step 1: Determine the initial velocities in the x and y directions.

The initial velocity (v0) is given as 25.0 m/s at an angle of 40.0°. We can use trigonometry to find the initial velocities in the x and y directions.

v0x = v0 * cos(40.0°) = 25.0 m/s * cos(40.0°) = 19.15 m/s v0y = v0 * sin(40.0°) = 25.0 m/s * sin(40.0°) = 16.07 m/s

Step 2: Determine the time it takes for the stone to hit the ground.

The time (t) is given as 3.75 s.

Step 3: Determine the horizontal distance the stone travels.

The horizontal distance (d) can be found using the equation d = v0x * t.

d = 19.15 m/s * 3.75 s = 71.81 m

Therefore, the stone lands approximately 71.81 meters from the foot of the cliff.