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.