an object is dropped from rest the top of a building 19.6 m hi. calculate the a) time for the object to reach the ground and b) object's speed just before it strikes the ground

h=1/2 g*t^2

time=sqrt(2*19.6m/(9.8m/s^2)
=2 seconds

vf=g*t=9.8*2 m/s

To calculate the time it takes for the object to reach the ground, you can use the equation of motion:

h = (1/2) * g * t^2

Where:
h is the height of the building (19.6 m)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken for the object to reach the ground.

First, let's rearrange the equation to solve for t:

t^2 = (2 * h) / g

To find the time, take the square root of both sides:

t = √((2 * h) / g)

Now, substitute the given value for h into the equation:

t = √((2 * 19.6) / 9.8)
= √(39.2 / 9.8)
= √4
= 2 seconds

Therefore, the object takes 2 seconds to reach the ground.

To calculate the object's speed just before it strikes the ground, you can use the equation:

v = g * t

Where:
v is the final velocity of the object just before it strikes the ground.

Substituting the values we have:

v = 9.8 * 2
= 19.6 m/s

So, the object's speed just before it strikes the ground is 19.6 m/s.