the distance from the base of the cliff to where the rock hits the level ground below the cliff is most nearly: C) 20 m.
Given the following data:
We know that acceleration due to gravity (a) for an object in free fall is equal to 9.8 meter per seconds square.
To find the distance from the base of the cliff to where the rock hits the level ground below the cliff:
First of all, we would determine the time by using the second equation of motion.
Mathematically, the second equation of motion is given by the formula;
[tex]S = ut + \frac{1}{2} at^2[/tex]
Where:
Substituting the values into the formula, we have;
[tex]20 = 0t + \frac{1}{2} (9.8)t^2\\\\20 = 0t + 4.9t^2\\\\20 = 4.9t^2\\\\t^2 = \frac{20}{4.9} \\\\t = \sqrt{4.082}[/tex]
Time, t = 2.02 seconds
Now, we can find the horizontal distance:
[tex]Horizontal\; distance = horizontal\; speed[/tex] × [tex]time[/tex]
[tex]Horizontal\; distance = 10[/tex] × [tex]2.02[/tex]
Horizontal distance = 20.0 meters.
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The horizontal distance covered by the rock is 20 m
Given data:
The height of cliff is, h = 20 m.
The initial speed of throw is, u = 10 m/s.
Applying the second kinematic equation of motion as,
[tex]h=u't+\dfrac{1}{2}gt^{2} \\20=0 \times t+\dfrac{1}{2} \times 9.8t^{2} \\20=4.9t^{2}\\t=2.02 \;\rm s[/tex]
Now, horizontal distance covered by the rock is,
[tex]d=u \times t\\d = 10 \times 2.02\\d \approx 20 \;\rm m[/tex]
Thus, the horizontal distance covered by the rock is 20 m.
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