A rock is thrown upward with a velocity of 21 meters per second from the top of a 48 meter high cliff, and it misses the cliff on the way back down. When will the rock be 7 meters from the water, below? Round your answer to two decimal places.

To solve this problem, we can use the equations of motion for an object in free fall. Let's break down the problem step by step.

Step 1: Solve for the time it takes for the rock to reach its highest point (when it is at 48 meters above the cliff).

We know the initial velocity of the rock when it is thrown upward is 21 m/s, and the acceleration due to gravity is approximately 9.8 m/s² (ignoring air resistance). At its highest point, the rock's vertical velocity will be zero.
We can use the kinematic equation:
vf = vi + at

Where:
vf is the final velocity (zero at the highest point)
vi is the initial velocity (21 m/s)
a is the acceleration due to gravity (-9.8 m/s²)
t is the time it takes to reach the highest point (what we want to find)

Setting the final velocity to zero and solving for t:
0 = 21 - 9.8t
9.8t = 21
t = 21/9.8
t ≈ 2.14 seconds

So it takes approximately 2.14 seconds for the rock to reach its highest point.

Step 2: Calculate the total time of flight for the rock.

Since the rock is thrown upward and misses the cliff on the way down, it will take the same amount of time to reach the highest point as it does to fall back down to the same height below the cliff.
Therefore, the total time of flight is twice the time it takes to reach the highest point:
Total time of flight = 2 * 2.14 seconds
Total time of flight ≈ 4.28 seconds

Step 3: Determine the position of the rock when it is 7 meters below the cliff.

The position of the rock can be calculated using the equation of motion:
s = vi * t + (1/2) * a * t²

Where:
s is the displacement (what we want to find)
vi is the initial velocity (21 m/s)
t is the time it takes for the rock to reach the desired position (unknown)
a is the acceleration due to gravity (-9.8 m/s²)

Rearranging the equation and plugging in the given values:
s = 21 * t + (1/2) * (-9.8) * t²
s = 21t - 4.9t²

Since we want the rock to be 7 meters below the cliff, we set s = -7:
-7 = 21t - 4.9t²

This equation can be rearranged into a quadratic equation:
4.9t² - 21t - 7 = 0

We can use the quadratic formula to solve for t:
t = (-b ± √(b² - 4ac)) / (2a)

Where:
a = 4.9
b = -21
c = -7

Substituting these values into the quadratic formula and solving for t, we get the two roots of the equation: t ≈ 1.26 seconds and t ≈ 3.01 seconds.

However, since the question asks for when the rock will be 7 meters below the cliff, we need to consider only the positive value of t, which is t ≈ 3.01 seconds.

Therefore, the rock will be 7 meters from the water below approximately 3.01 seconds after it was thrown upward.