How long would it take for a ball dropped from the top of a 256 foot building to hit the ground? Round your answer to two decimal places
Since it is "dropped", there is no initial velocity, so
distance = -16t^2 + 256 , where t is in seconds and distance in feet
when it hits the ground, distance = 0
0 = -16t^2 + 256
16t^2 = 256
t^2 = 16
t = .....
4 seconds
To determine how long it would take for a ball to hit the ground after being dropped from a 256-foot building, we can use the equation for free fall:
d = 1/2 * g * t^2
Where:
- d is the distance traveled (256 feet in this case)
- g is the acceleration due to gravity (approximately 32.2 feet/second^2)
- t is the time in seconds
Rearranging the equation to solve for time, we get:
t = sqrt(2d / g)
Now, let's plug in the given values and calculate the time:
t = sqrt(2 * 256 / 32.2)
t = sqrt(512 / 32.2)
t = sqrt(15.9)
t ≈ 3.99 seconds
Therefore, it would take approximately 3.99 seconds for the ball to hit the ground when dropped from the top of a 256-foot building. Rounded to two decimal places, the answer is 4.00 seconds.
To calculate the time it takes for a ball to fall from the top of a building, we can use the kinematic equation:
h = 0.5 * g * t^2
Where:
h = height (in this case, 256 ft)
g = acceleration due to gravity (32.2 ft/s^2)
Rearranging the equation to solve for time (t):
t = sqrt(2h/g)
Now we can substitute the values and calculate:
t = sqrt(2 * 256 ft / 32.2 ft/s^2)
t = sqrt(512/32.2)
t = sqrt(15.9046)
t ≈ 3.99 seconds (rounded to two decimal places)
Therefore, it would take approximately 3.99 seconds for the ball to hit the ground when dropped from the top of the 256-foot building.