The Mars rover drops a rock off the rim of a 120 meter high cliff on Mars. The free-fall acceleration is -3.7m/s2. What is the velocity that the rock hits the ground?
as on earth
v^2 = 2as = 2(3.7)(120)
v = -27.8 m/s
Note that you kind of have to fudge on the negative sign to take the square root, then apply it again at the end.
Well, it sounds like the Mars rover has a bit of gravity trouble there. But don't worry, I'm here to help! To find the velocity at which the rock hits the ground, we can use the kinematic equation for free-fall motion:
v² = u² + 2as
Where:
v is the final velocity (what we're trying to find)
u is the initial velocity (which is zero since the rock was dropped)
a is the acceleration due to gravity (-3.7m/s²)
s is the vertical distance (120 meters)
Plugging in the values, we have:
v² = 0 + 2(-3.7)(120)
v² = -888
v ≈ √-888
Now, the square root of a negative number doesn't exist in the realm of real numbers. So, unless we start venturing into the imaginary realm, it looks like the answer is... wait for it... "rock-bottom." That's right, the velocity at which the rock hits the ground on Mars is zero. Of course, in reality, the rock would actually have a non-zero final velocity due to air resistance, but for the purposes of this question, we're assuming a perfect vacuum. Now, that's a gravity-gone-hilarious situation!
To calculate the velocity at which the rock hits the ground, we can use the equations of motion for free-fall:
v^2 = u^2 + 2as
where:
v is the final velocity (what we want to find),
u is the initial velocity (which is 0 in this case, as the rock is dropped),
a is the acceleration due to gravity (-3.7 m/s^2),
s is the vertical displacement (in this case, the height of the cliff, 120 meters).
Substituting the given values into the equation:
v^2 = 0 + 2(-3.7)(-120)
v^2 = 2(3.7)(120)
v^2 = 888
Taking the square root of both sides to solve for v:
v = sqrt(888)
v ≈ 29.8 m/s
Therefore, the velocity at which the rock hits the ground is approximately 29.8 m/s.
To find the velocity at which the rock hits the ground, we can use the equations of motion.
First, we need to determine the time it takes for the rock to fall from the top of the cliff to the ground. We can use the equation:
y = ut + (1/2)at^2
Where:
y = change in height (the height of the cliff) = -120 meters (negative value because the rock is falling)
u = initial velocity = 0 m/s (the rock starts from rest)
a = acceleration = -3.7 m/s^2 (negative value because it is directed downwards)
t = time taken (unknown)
Rearranging the equation, we have:
-120 = 0 * t + (1/2) * (-3.7) * t^2
-120 = -1.85t^2
Dividing both sides by -1.85:
t^2 = 120 / 1.85
t^2 ≈ 64.8649
Taking the square root of both sides, we find:
t ≈ √(64.8649)
t ≈ 8.06 seconds (rounded to two decimal places)
Now that we know the time it takes for the rock to fall, we can find the velocity at which it hits the ground using the equation:
v = u + at
Where:
v = final velocity (the velocity at which the rock hits the ground, which we want to find)
u = initial velocity = 0 m/s (starting from rest)
a = acceleration = -3.7 m/s^2 (negative value because it is directed downwards)
t = time taken = 8.06 seconds (rounded to two decimal places)
Substituting the given values into the equation, we have:
v = 0 + (-3.7) * 8.06
v ≈ -29.86 m/s (rounded to two decimal places)
Therefore, the velocity at which the rock hits the ground is approximately -29.86 m/s (negative value because it is directed downwards).