a rock is dropped off a cliff. after 4s, what is the rocks velocity? how far did the rock fall?
1. Vf = Vo + gt,
Vf = 0 + 9.8*4 = 39.2m/s.
2. d = Vo*t + 0.5g*t^2,
d = 0 + 4.9*4^2 = 78.4m.
To calculate the rock's velocity after 4 seconds and the distance it fell, we need to use the equations of motion.
The equation for calculating the velocity of an object in free fall is:
v = u + gt
Where:
- v is the final velocity
- u is the initial velocity (in this case, it is 0 because the rock is dropped from rest)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time
Plugging in the values into the equation, we have:
v = 0 + (9.8 m/s^2)(4 s)
v = 39.2 m/s
Therefore, the rock's velocity after 4 seconds of falling is 39.2 m/s.
Now, to calculate the distance the rock fell, we use the equation for displacement:
s = ut + (1/2)gt^2
Where:
- s is the displacement or distance
- u is the initial velocity (0 m/s)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time
Substituting the values into the equation:
s = (0 m/s)(4 s) + (1/2)(9.8 m/s^2)(4 s)^2
s = 0 + (1/2)(9.8 m/s^2)(16 s^2)
s = 0 + 78.4 m
Therefore, the rock fell a distance of 78.4 meters after 4 seconds.