A ball falls from a height of 18m above the ground. Find the speed with which the ball hits the ground. \([g = 10ms^{-2}]\)
Answer Details
We can use the formula for the final velocity of an object falling from a height under the influence of gravity:
$v^2 = u^2 + 2gh$
where $v$ is the final velocity, $u$ is the initial velocity (in this case, 0), $g$ is the acceleration due to gravity and $h$ is the height from which the object is falling.
Substituting the given values, we have:
$v^2 = 0 + 2\times10\times18 = 360$
Taking the square root of both sides, we get:
$v = \sqrt{360} = 6\sqrt{10} \approx 18.97ms^{-1}$
Therefore, the speed with which the ball hits the ground is approximately 18.97\(ms^{-1}\). Hence, the correct option is (d) 18.97\(ms^{-1}\).