A rock is thrown horizontally off of a cliff. It starts with a velocity of 4.2 m/s and the cliff is
68.2 m high.
How long does the rock fall?
I got 3.7 seconds but I do not know if it is right.
(Help on these two)
How far out from the bottom of the cliff does the rock fall?
What is the rock’s velocity just before it hits the ground?
v^2=u^2-2gt
0=4.2^2-2*10t
20gt=17.64
t=0.8sec
v=u+gt
v=4.2*10*0.8
v=33.6m/s
assume g to be 10ms-2 because it makes calculation much easier
To find the time it takes for the rock to fall, we can use the equation of motion for free-falling objects:
h = (1/2) * g * t^2
Where h is the height of the cliff (68.2 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
We can rearrange the equation to solve for t:
t = sqrt(2h/g)
Plugging in the values:
t = sqrt((2 * 68.2 m) / (9.8 m/s^2))
t ≈ 3.70 s
So your answer of 3.7 seconds is correct for the time it takes for the rock to fall.
To find how far out from the bottom of the cliff the rock falls, we can use the formula:
d = v * t
Where d is the distance traveled by the rock, v is the horizontal velocity of the rock, and t is the time.
Since the rock is thrown horizontally, it will maintain the same horizontal velocity (4.2 m/s) throughout its motion. Therefore:
d = (4.2 m/s) * (3.70 s)
d ≈ 15.54 m
So, the distance out from the bottom of the cliff where the rock falls is approximately 15.54 meters.
Finally, to find the rock's velocity just before it hits the ground, we can use the equation:
v_final = v_initial + g * t
Where v_final is the final velocity, v_initial is the horizontal velocity (4.2 m/s), g is the acceleration due to gravity (9.8 m/s^2), and t is the time (3.7 s).
Plugging in the values:
v_final = 4.2 m/s + (9.8 m/s^2) * 3.70 s
v_final ≈ 42.14 m/s
Therefore, the rock's velocity just before it hits the ground is approximately 42.14 m/s.