Suppose that an object is dropped from a height of h meters and hits the ground with a velocity of 1 meters per second. Then = √19.6h. If an object hits the ground with a velocity of 20.4 meters per second, from what height was it dropped?
From the given information, we can equate the final velocity (vf) with the initial velocity (u) plus the product of acceleration (a) and time (t):
vf = u + at
Since the object is dropped, the initial velocity (u) is equal to 0, so the equation simplifies to:
vf = at
Now, we can use the equation for velocity (v) in terms of acceleration (a) and distance (d):
v^2 = u^2 + 2ad
Since the object is dropped, the initial velocity (u) is equal to 0, so the equation further simplifies to:
v^2 = 2ad
Rearranging the equation and solving for distance (d):
d = v^2 / (2a)
Substituting the value of acceleration (a = 9.8 m/s^2), we get:
d = v^2 / (2 * 9.8)
For the object hitting the ground with a velocity of 1 m/s:
d = 1^2 / (2 * 9.8) = 0.051 meters
For the object hitting the ground with a velocity of 20.4 m/s:
d = 20.4^2 / (2 * 9.8) = 42 meters
Therefore, the object was dropped from a height of 42 meters.