If an object is thrown in an upward direction from the top of a building 1.60 x 102 ft. high at an initial velocity of 21.82 mi/h, what is its final velocity when it hits the ground? (Disregard wind resistance. Round answer to nearest whole number and do not reflect negative direction in your answer.)
I guessed 96 ft/s, would someone please show me what equation to use as well as the answer so I can understand how to do it? Please and thank you
You guessed? That is a silly thing to do.
vf^2=vi^2+2ah
a=-9.8m/s^2 h=-1.60E2
To solve this problem, we need to use the kinematic equations of motion. Specifically, we can use the following equation:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity
- vi is the initial velocity
- a is the acceleration
- d is the distance traveled
In this case, we want to find the final velocity when the object hits the ground, so d is the height of the building. The initial velocity is given as 21.82 mi/h, but we need to convert it to feet per second to ensure consistent units. To do this, we can use the conversion factor that 1 mi = 5280 feet, and 1 hour = 3600 seconds:
vi = (21.82 mi/h) * (5280 ft/mi) * (1/3600 h/s)
vi ≈ 31.96 ft/s
The initial velocity is approximately 31.96 ft/s.
The acceleration can be determined using the equation:
vf^2 = vi^2 + 2ad
Since the object is thrown upward against gravity, the acceleration is negative (opposite direction to velocity):
a = -32.17 ft/s^2 (approximately, considering acceleration due to gravity)
Now, we can substitute the known values into the equation and solve for vf:
vf^2 = (31.96 ft/s)^2 + 2*(-32.17 ft/s^2)(-1.6 * 10^2 ft)
vf^2 = 1023.04 ft^2/s^2 + 102.72 ft^2/s^2
vf^2 ≈ 1125.76 ft^2/s^2
Finally, we take the square root of both sides of the equation to find vf:
vf ≈ √1125.76 ft^2/s^2
vf ≈ 33.55 ft/s
Rounding this to the nearest whole number: vf ≈ 34 ft/s
Therefore, the final velocity of the object, when it hits the ground, is approximately 34 ft/s.