Two rocks are thrown off the edge of a cliff that is 15.0 m above the ground. The first rock is thrown upward, at a velocity of +12.0 m/s. The second is thrown downward, at a velocity of −12.0 m/s. Ignore air resistance.
Determine (a) how long it takes the first rock to hit the ground and (b) at what velocity it hits. Determine (c) how long it takes the second rock to hit the ground and (b) at what velocity it hits.
Both are solved the same way:
Hf=hi+vi*t-4.9t^2
put in hi (15), hf(zero), vi (either 12, or-12), and solve for t.
velocity? Vf=Vi-9.8t
To solve this problem, we can use the kinematic equations of motion. Let's break down each part of the problem:
(a) To determine how long it takes the first rock to hit the ground, we can use the equation:
h = ut + (1/2)at^2
where:
h = height (15.0 m)
u = initial velocity (+12.0 m/s)
a = acceleration (-9.8 m/s^2) due to gravity
t = time (unknown)
We need to rearrange the equation to solve for t:
t = sqrt(2h / g)
Substituting the given values into the equation:
t = sqrt(2 * 15.0 / 9.8)
Calculating this, we find t ≈ 1.77 seconds.
So, the first rock takes approximately 1.77 seconds to hit the ground.
(b) To determine the velocity at which the first rock hits the ground, we can use the equation:
v = u + at
where:
v = final velocity (unknown)
u = initial velocity (+12.0 m/s)
a = acceleration (-9.8 m/s^2) due to gravity
t = time (1.77 seconds)
Substituting the values into the equation:
v = 12.0 - 9.8 * 1.77
Calculating this, we find v ≈ -16.6 m/s.
So, the first rock hits the ground with a velocity of approximately -16.6 m/s (negative sign indicates the direction downward).
Now, let's move on to the second rock:
(c) To determine how long it takes the second rock to hit the ground, we can use the same equation as before:
t = sqrt(2h / g)
Substituting the given values:
t = sqrt(2 * 15.0 / 9.8)
Calculating this, we again find t ≈ 1.77 seconds.
Therefore, the second rock also takes approximately 1.77 seconds to hit the ground.
(d) To determine the velocity at which the second rock hits the ground, we can use the same equation as before:
v = u + at
where:
v = final velocity (unknown)
u = initial velocity (-12.0 m/s)
a = acceleration (-9.8 m/s^2) due to gravity
t = time (1.77 seconds)
Substituting the values into the equation:
v = -12.0 - 9.8 * 1.77
Calculating this, we find v ≈ -30.4 m/s.
So, the second rock hits the ground with a velocity of approximately -30.4 m/s (negative sign indicates the direction downward).