a rock is dropped off a very high cliff and reaches the ground 6.00s later
how fast was the rock traveling just before it hit the ground
To find the speed of the rock just before it hit the ground, we can use the kinematic equation:
v = u + at
where:
v = final velocity (speed just before hitting the ground)
u = initial velocity (initial speed at the top of the cliff, which is 0 m/s)
a = acceleration due to gravity (9.81 m/s^2)
t = time taken to reach the ground (6.00 s)
Plugging in the values:
v = 0 + (9.81 m/s^2)(6.00 s)
v = 0 + 58.86
v = 58.86 m/s
Therefore, the speed of the rock just before it hit the ground was 58.86 m/s.