Question

While visiting the Sky Tower you happen to drop some coins from the top floor which is 222 m above the ground below. Assuming that the coins do not reach terminal velocity and accelerate at 9.81 m/s2 , what is the speed just before they reach the ground? Write you answer in m/s2

Answer 1

Answer on Question #69588, Physics / Other

While visiting the Sky Tower you happen to drop some coins from the top floor which is 222 m above the ground below. Assuming that the coins do not reach terminal velocity and accelerate at $9.81\mathrm{m / s^2}$, what is the speed just before they reach the ground? Write you answer in $\mathrm{m / s^2}$.

Solution:

The kinematic equation that describes an object's motion is:

$$v _ {f} ^ {2} = v _ {i} ^ {2} + 2 a d$$

The symbol $d$ stands for the displacement of the object. The symbol $a$ stands for the acceleration of the object. And the symbol $v$ stands for the velocity of the object; a subscript of $i$ after the $v$ indicates that the velocity value is the initial velocity value and a subscript of $f$ indicates that the velocity value is the final velocity value.

In our case

$$v _ {i} = 0 \mathrm{~m / s}$$ $$d = 222 \mathrm{~m}$$ $$a = 9.81 \mathrm{~m / s^2}$$ $$v _ {f} = \sqrt {2 a d} = \sqrt {2 \times 9.81 \times 222} = 66.0 \mathrm{~m / s}$$

Answer: $66.0 \, \text{m/s}$

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