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.
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
1 answer
1 answer