A student atands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above flat, horizontal beach. With what speed of impact does the stone land?
What is the initial KE? What is the initial PE?
FinalKE=intialPE+initialKE
how does that help mee???
To find the speed of impact at which the stone lands, we can use the concept of projectile motion. Since the stone is thrown horizontally over the edge of the cliff, its initial vertical velocity is zero. The only force acting on the stone in the vertical direction is gravity.
To solve this problem, we can use the following equations of motion:
1. Vertical displacement:
h = ut + (1/2)gt^2
2. Initial vertical velocity:
u = 0 (since the stone is thrown horizontally)
3. Acceleration due to gravity:
g = 9.8 m/s^2 (assuming no air resistance)
We are given that the vertical displacement (h) is 50.0 m, and we need to find the time it takes for the stone to hit the ground.
Using equation (1), we can rewrite it as:
50 = (1/2) * 9.8 * t^2
Simplifying the equation, we get:
t^2 = 50 / (1/2 * 9.8)
t^2 = 10.2041
Taking the square root of both sides, we find that:
t ≈ 3.19 s (rounded to two decimal places)
Now, to find the speed of impact, we can use the equation:
v = u + gt
Since the initial vertical velocity (u) is 0, the equation simplifies to:
v = gt
Substituting the values, we get:
v ≈ 9.8 * 3.19
v ≈ 31.26 m/s
Therefore, the stone lands with a speed of approximately 31.26 m/s.