A stone is dropped from a cliff. Its speed after it has fallen 100 m ...
Explanation:
When an object is dropped from a height, it falls freely under the influence of gravity. The acceleration due to gravity is approximately 9.8 m/s^2. So, as the stone falls, its speed increases due to the acceleration.
Using the kinematic equation:
We can use the kinematic equation to calculate the final speed of the stone after it has fallen 100 m. The equation is:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (0 m/s, as the stone is dropped)
a = acceleration due to gravity (9.8 m/s^2)
s = displacement (100 m)
Substituting the given values:
v^2 = 0^2 + 2 * 9.8 * 100
v^2 = 0 + 1960
v^2 = 1960
Taking the square root of both sides:
v = √1960
v ≈ 44.2 m/s
Therefore, the speed of the stone after it has fallen 100 m is approximately 44.2 m/s. So, option B is the correct answer.
A stone is dropped from a cliff. Its speed after it has fallen 100 m ...
In this case we have as, u = 0m/s, v=?, h= 100m, g=9.8m/s^2
now , by using 3rd law of motion which is
v^2 -u^2 =2gh
V^2 - 0 = 2×9.8 ×100
v^2 = 1960
v=√ 1960 = 44.27 m/s