A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of 13 m/s. The cliff 42 m above the water’s surface With what speed does the stone strike the water? Answer in units of m/s..
To determine the speed at which the stone strikes the water, we can make use of the principles of projectile motion.
The key here is to recognize that the stone will follow a parabolic trajectory after leaving the person's foot until it hits the water.
Let's break down the problem into horizontal and vertical components:
1. Vertical motion:
The stone is subjected to the acceleration due to gravity, which is approximately 9.8 m/s^2. The initial vertical velocity is 0 m/s (as the stone is kicked horizontally) and it travels a vertical distance of 42 m.
We can use the equation of motion:
s = ut + 1/2 * a * t^2
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values:
42 = 0 * t + 1/2 * 9.8 * t^2
Rearranging the equation:
4.9t^2 = 42
Simplifying:
t^2 = 42 / 4.9
t^2 ≈ 8.57
Taking the square root of both sides:
t ≈ √8.57
t ≈ 2.93 seconds
2. Horizontal motion:
The stone is kicked horizontally, which means its horizontal velocity remains constant throughout its motion. Therefore, the horizontal component of its velocity is 13 m/s.
3. Combining horizontal and vertical components:
The speed at which the stone strikes the water is the resultant of its horizontal and vertical velocities, which can be obtained using the Pythagorean theorem.
v = √(vx^2 + vy^2)
vx = 13 m/s (horizontal component)
vy = -9.8 m/s^2 * 2.93 s (vertical component, taking acceleration due to gravity and time into account)
Substituting the values:
v = √(13^2 + (-9.8 * 2.93)^2)
v ≈ √(169 + (-84.864)^2)
v ≈ √(169 + 7,191.334) ≈ √7360.334
v ≈ 85.79 m/s (rounded to two decimal places)
Therefore, the speed at which the stone strikes the water is approximately 85.79 m/s.