Final answer:
The position of a dropped rock at any time t is given by the equation s(t) = h - 4.9t² for an initial height of 19.5 meters. The velocity of the rock at t = 1/2 second is 4.9 m/s. These calculations assume negligible air resistance and use the standard acceleration due to gravity.
Step-by-step explanation:
To answer the student's question, we'll use the standard equation of motion for an object under the influence of gravity when air resistance is negligible. The position s(t) of the rock at any time t after being dropped from rest from a height h is given by the equation:
s(t) = h - ½gt²
Where:
- h is the initial height.
- ½g is the acceleration due to gravity (½ of 9.8 m/s²).
- t is the time in seconds since the rock was dropped.
For a 64 foot (approximately 19.5 meters) platform, this becomes:
s(t) = 19.5 - 4.9t²
To find out how fast the rock is traveling at t = 1/2 second:
We use the velocity equation of a falling object v(t) = gt.
Thus, the velocity v(1/2) at t = 1/2 second is:
v(1/2) = 9.8 * (1/2) = 4.9 m/s
The rock is traveling at a velocity of 4.9 meters per second at t = 1/2 second.