A stone is dropped from a height of 144 feet.
a. When will the stone hit the ground?
b. With what velocity does it hit the ground?
Please disregard. I solved the problem. Thank you anyways.
To find the answers to these questions, we can use the equations of motion for vertically falling objects. The key equation we will be using is:
𝑑 = 𝑣0𝑡 + 0.5𝑎𝑡^2
Where:
𝑑 is the displacement (height or distance covered),
𝑣0 is the initial velocity (in this case, 0 as the stone is dropped),
𝑡 is the time, and
𝑎 is the acceleration due to gravity (approximately -32.17 ft/s^2 near the Earth's surface).
a. When will the stone hit the ground?
To find the time it takes for the stone to hit the ground, we need to solve the equation 𝑑 = 0, as the displacement will be zero when it hits the ground. Plugging in the values:
0 = 0(𝑡) + 0.5(-32.17𝑡^2)
Simplifying the equation further:
-16.085𝑡^2 = 0
Since this is a quadratic equation (with t^2 term), we can solve it by setting each factor equal to zero:
𝑡 = 0 or 𝑡 = 0
Since time cannot be zero (as the stone has been dropped from a height), the only solution is 𝑡 = 0. This means that the stone hits the ground immediately when it is dropped.
b. With what velocity does it hit the ground?
To find the velocity at which the stone hits the ground, we will use the equation:
𝑣 = 𝑣0 + 𝑎𝑡
Plugging in the values:
𝑣 = 0 + (-32.17𝑡)
Again, since 𝑡 = 0, the velocity equation becomes:
𝑣 = 0 + (-32.17 * 0)
Simplifying, we get:
𝑣 = 0
This means that the stone hits the ground with a velocity of 0 ft/s.