If an object is thrown in an upward direction from the top of a building 160 ft. high at an initial speed of 21.82 mi/h., what is its final speed when it hits the ground?(Disregard wind resistance. Round answer according to significant figures and do not reflect negative direction in your answer.
the answer you gave is very incorrect according to the answer choices given
To find the final speed of the object when it hits the ground, we need to consider the concept of projectile motion. The object is thrown upward with an initial speed of 21.82 mi/h from a height of 160 ft.
First, we need to convert the initial speed from miles per hour to feet per second. There are 5280 feet in a mile, and 3600 seconds in an hour, so to convert 21.82 mi/h to ft/s:
21.82 mi/h * (5280 ft/mi) / (3600 s/h) ≈ 31.99 ft/s (rounded to two decimal places)
Next, we can use the fact that the change in potential energy of the object as it falls from the top of the building to the ground is equal to the change in its kinetic energy. In this case, the change in potential energy is given by the decrease in height (160 ft), and the initial kinetic energy is given by the initial speed (31.99 ft/s).
Using the formula for potential energy (PE = mgh) and kinetic energy (KE = 0.5mv^2), where m is the mass of the object, g is the acceleration due to gravity (32.17 ft/s^2), and h is the height:
mgh = 0.5mv^2
We can cancel out the mass and solve for v^2:
gh = 0.5v^2
v^2 = 2gh
v = √(2gh)
Plugging in the values we know:
v = √(2 * 32.17 ft/s^2 * 160 ft)
v ≈ √(10307.2 ft^2/s^2)
v ≈ 101.51 ft/s (rounded to two decimal places)
Therefore, the final speed of the object when it hits the ground is approximately 101.51 ft/s.
Kinetic energy increase = Potential energy decrease
(M/2) [V2^2 - V1^2] = M g H
[V2^2 - V1^2] = 2 g H
Solve for V2, the final velocity.
V1 = 21.82 m/s
The V2 answer will not depend upon the direction it was thrown!