A baseball is released at rest from the top of the washington Monument. It hits the ground after falling for 6.0 s. What was the height from which the ball was dropped? (g=9.8 m/s2 and assume air resistance is negligible)
h=1/2 g t^2
y = (1/2)(-9.80 m/s2)(6.0s)^2
y = (-4.9 m/s2)(36.0s2)
y = 176.4 m
The distance it falls in time t is
Y = (1/2) g t^2.
That is a formula you need to learn.
Apply it.
To find the height from which the baseball was dropped, we can use the formula for the distance fallen by an object in free fall:
d = 1/2 * g * t^2
Where:
d is the distance fallen
g is the acceleration due to gravity (9.8 m/s^2)
t is the time taken to fall (6.0 seconds)
In this case, we want to find the height from which the ball was dropped, so we can rearrange the formula to solve for d:
d = 1/2 * g * t^2
d = 1/2 * 9.8 * (6.0)^2
d = 1/2 * 9.8 * 36.0
d = 176.4 meters
Therefore, the height from which the ball was dropped is 176.4 meters.