A stone on the edge of a vertical cliff is catapulted horizontally with a speed of 4.0 m/s. If the cliff is 0.20 km high, calculate the time taken (in seconds) for the stone to reach the ground.Assume g = 9.8 m/s2; and give your answer with the appropriate number of significant figures.
Question
A stone on the edge of a vertical cliff is catapulted horizontally with a speed of 4.0 m/s. If the cliff is 0.20 km high, calculate the time taken (in seconds) for the stone to reach the ground.Assume g = 9.8 m/s2; and give your answer with the appropriate number of significant figures.
Solution
The problem involves a stone being projected horizontally from a cliff. The time it takes for the stone to hit the ground is only dependent on the vertical motion. So, we can treat this as a one-dimensional motion problem in the vertical direction.
The height of the cliff is given as 0.20 km, which is 200 m (since 1 km = 1000 m).
We know that the equation for distance travelled under constant acceleration is:
d = vit + 0.5at^2
where: d is the distance (200 m in this case), vi is the initial velocity (0 m/s in this case, because the stone is initially not moving vertically), a is the acceleration (9.8 m/s^2, which is the acceleration due to gravity), and t is the time we want to find.
Substituting the known values into the equation, we get:
200 m = 0t + 0.59.8 m/s^2 * t^2
Solving this equation for t gives:
t^2 = 200 m / (0.5*9.8 m/s^2) = 40.82 s^2
Taking the square root of both sides gives:
t = sqrt(40.82 s^2) = 6.39 s
So, the stone will take approximately 6.4 seconds to hit the ground, to three significant figures.
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