A rock is dropped from a height of 60 m and is in free fall. What is the velocity of the rock as it reaches the ground 3.5 seconds later?
20 m/s downward
v=gt=9.8 x 3.5 =34.3 m/s
To find the velocity of the rock as it reaches the ground, we can use the equation of motion for an object in free fall:
v = u + at
Where:
v = final velocity
u = initial velocity (which is 0 since the rock is dropped)
a = acceleration due to gravity (approximately -9.8 m/s^2)
t = time
In this case, the rock is dropped from a height of 60 m, so we need to find the time it takes for the rock to reach the ground. We can use the equation for free fall:
h = ut + (1/2)at^2
where:
h = height
u = initial velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time
We rearrange the equation to solve for time:
h = (1/2)at^2
2h = -9.8t^2
t^2 = (2h) / 9.8
t = square root of [(2h) / 9.8]
Substituting the given values:
t = square root of [(2 * 60) / 9.8]
t ≈ 3.83 seconds
Now that we have the time it takes for the rock to reach the ground, we can find the velocity using the first equation:
v = u + at
v = 0 + (-9.8 * 3.83)
v ≈ -37.87 m/s (rounded to two decimal places)
Therefore, the velocity of the rock as it reaches the ground is approximately -37.87 m/s (since upward is conventionally considered positive and downward is negative in this scenario).
Vf=Vi+gt
Vf= final velocity (what we're solving for)
Vi= initial velocity (in our problem, initial velocity is zero because the rock starts from rest)
g=gravitational acceleration (9.8m/s/s)
t= time (in our case, 3.5s)
Vf= 0m/s + (9.8m/s/s)(3.5s)
Vf= 34.3m/s downward (you must have the direction because velocity is a vector quantitiy, meaning that it is represented by both magnitude and direction)
(source AP physics B)