A stone is dropped from a height of 45m. Find the time it takes to hit the ground. \([g = 10 ms^{-2}]\)
Answer Details
To solve this problem, we can use the formula:
\[
s = ut + \frac{1}{2}at^2
\]
where s is the distance, u is the initial velocity, t is the time, and a is the acceleration due to gravity. In this case, the initial velocity is zero because the stone is dropped from rest. We also know that the distance is 45m and the acceleration due to gravity is 10 \(ms^{-2}\).
Thus, we have:
\begin{align*}
s &= ut + \frac{1}{2}at^2 \\
45 &= 0t + \frac{1}{2}(10)t^2 \\
45 &= 5t^2 \\
t^2 &= 9 \\
t &= 3
\end{align*}
Therefore, the time it takes for the stone to hit the ground is 3 seconds.
Hence, the answer is (a) 3.0 seconds.