A robot probe drops a camera off the rim of a 457 m high cliff on Mars, where the free-fall acceleration is 3.7 m/s2. find the time required for the camera to reach the ground. answer in units of s.

i don't know if my equation is right:
* tfinal= (vinitial + g)(vfinal)

To find the time required for the camera to reach the ground, we can use the equation for motion under constant acceleration:

d = v_initial * t + (1/2) * a * t^2

Where:
- d is the distance traveled (457 m)
- v_initial is the initial velocity (0 m/s, since the camera is dropped)
- a is the acceleration (3.7 m/s^2)
- t is the time we want to find.

To find the time, we need to rearrange the equation to solve for t. Start by substituting the values:

457 m = 0 m/s * t + (1/2) * (-3.7 m/s^2) * t^2

Simplifying:

457 m = -1.85 m/s^2 * t^2

Divide both sides by -1.85 m/s^2:

-247.03 = t^2

Take the square root of both sides:

t = √(-247.03)

Here, we encounter a problem. The result is imaginary, which means there is no real solution. This indicates that the camera will not reach the ground. However, since the question asks for the time required, we can consider the maximum height the camera will reach before falling back down.

Using the equation for maximum height reached (when the final velocity is 0), we have:

v_final = v_initial + a * t
0 = 0 + 3.7 m/s^2 * t

Solving for t:

t = 0 / 3.7 m/s^2 = 0 s

Therefore, the time required for the camera to reach the ground is 0 seconds.