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 rockβs velocity just before it hits the ground?
Correction: Yf = 36.6 m/s.
h = 0.5g*t^2 = 68.2 m.
4.9t^2 = 68.2
t^2 = 13.92.
t = 3.73 s.
Dx = Xo*t = 4.2m/s * 3.73s. = 15.7 m.
Yf^2 = Yo^2 + 2g*h.
Yf^2 = 0 + 19.6*68.2 = 1336.72.
Yf = 36.6 m.
Vf = sqrt(Xo^2 + Yf^2) = Final velocity.
To calculate how long the rock falls, we can use the formula for the vertical motion:
π¦ = π£βπ‘ + 0.5ππ‘Β²
where:
π¦ is the height of the cliff (68.2 m),
π£β is the initial vertical velocity (0 m/s since it was thrown horizontally),
π‘ is the time, and
π is the acceleration due to gravity (-9.8 m/sΒ²).
Plugging in the values, we get:
68.2 = 0.5(-9.8)π‘Β²
Simplifying the equation, we get:
-4.9π‘Β² = 68.2
Dividing both sides by -4.9, we get:
π‘Β² = -14
Since time cannot be negative, it means that there is no real solution for π‘ in this equation. Therefore, the rock does not fall vertically and remains at the same height.
As for the horizontal distance the rock travels before hitting the ground, we can use the horizontal motion formula:
π = π£βππ‘
where:
π is the horizontal distance (what we need to find),
π£βπ is the horizontal velocity (4.2 m/s), and
π‘ is the time (which we cannot find using the information provided).
Unfortunately, since we don't know the time the rock takes to reach the ground, we cannot calculate the exact horizontal distance it travels.
Regarding the rock's velocity just before it hits the ground, we know that the horizontal velocity remains constant throughout its motion. Therefore, the rock's velocity just before hitting the ground will also be 4.2 m/s horizontally.
To find the time it takes for the rock to fall, you can use the equation of motion for vertical motion, which is h = (1/2)gt^2, where h is the height of the cliff, g is the acceleration due to gravity, and t is the time.
Given that the height of the cliff is 68.2 m, we can plug in the value of h into the equation:
68.2 = (1/2)(9.8)t^2
To solve for t, we can rearrange the equation:
t^2 = (68.2 Γ 2) / 9.8
t^2 = 13.94
Taking the square root of both sides yields:
t β 3.73 seconds
The time it takes for the rock to fall is approximately 3.73 seconds. Therefore, your answer of 3.7 seconds is close enough.
To determine how far out from the bottom of the cliff the rock falls, we need to calculate the horizontal distance traveled by the rock. Since the rock is thrown horizontally, its horizontal velocity remains constant throughout its motion.
The distance traveled horizontally is given by the equation d = v * t, where d is the distance, v is the horizontal velocity, and t is the time.
We are given that the rock has a horizontal velocity of 4.2 m/s and the time it takes for the rock to fall is 3.73 seconds. So, we can plug in these values into the equation:
d = 4.2 * 3.73
d β 15.666 meters
Therefore, the rock falls approximately 15.666 meters out from the bottom of the cliff.
Lastly, to find the rock's velocity just before it hits the ground, we can use the equation of motion:
v = gt
Since the rock is falling vertically, its velocity just before hitting the ground will be the same as the velocity it acquired during its free fall. Therefore, we can use the value of g (acceleration due to gravity) and the time it takes to fall:
v = 9.8 * 3.73
v β 36.554 m/s
Hence, the rock's velocity just before it hits the ground is approximately 36.554 m/s.