a baseball is released at rest from the top of a building.it hits the ground after falling for 6seconds . what was the height from which the ball was dropped
(1/2) a t^2 = h
(1/2)(9.81)(36) = h in meters
1.8*10^2m
To find the height from which the ball was dropped, we can use the formula for the distance fallen by an object in free fall:
d = (1/2)gt^2
Where:
d is the distance fallen
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time the object falls
In this case, the time the ball falls is given as 6 seconds. Substituting the values into the formula, we have:
d = (1/2)(9.8)(6)^2
d = (1/2)(9.8)(36)
d = 4.9(36)
d = 176.4 meters
Therefore, the height from which the ball was dropped is approximately 176.4 meters.
To determine the height from which the ball was dropped, we can use the formula for the distance traveled by an object in free fall:
d = (1/2) * g * t^2
Where:
d is the distance traveled
g is the acceleration due to gravity
t is the time.
We know that the ball fell for 6 seconds and the acceleration due to gravity is approximately 9.8 m/s^2.
Plugging in the values:
d = (1/2) * 9.8 * 6^2
d = 0.5 * 9.8 * 36
d = 176.4 meters
Therefore, the height from which the ball was dropped is approximately 176.4 meters.