Question

A 10 kg rock is suspended 320 m above the ground. Calculate the approximate speed with which the rock hits the ground.

Answer 1

Answer:

B. 79 m/s

Explanation:

Since we have information about the distance of the fall, we can use the following formula to find the final velocity of the object:

$v_{f}^2=v_{i}^2+2gh$

clearing for $v_{f}$

$v_{f}=\sqrt{v_{i}^2+2gh}$

where $v_ {f}$ is the final velocity (when it hits the ground), $v_ {i}$ the initial velocity, $g$ is the acceleration of gravity ($g=9.8m/s^2$), and $h$ is the height from where the rock is dropped, in this case: $h=320m$

Assuming that the rock was only dropped without adding an initial velocity, $v_ {i}$ is zero.

Substituting all the values in the formula:

$v_{f}=\sqrt{(0)^2+2(9.8m/s^2)(320m)}$

$v_{f}=\sqrt{6272}$

$v_{f}=79.2m/s$

The value that best approximates from the options is 79 m/s,  the answer is B.

Answer 2

The answer for this is actually b. 79 m/s

I submitted it to odyssey-ware and i got the question correct