a person standing at the edge of a cliff throws a rock directly upward. it observed that 2 seconds later, the rock is it maximum height and that 5 seconds after it reached the highest point, it hits the ground at the base of the cliff. what is the initial velocity of the rock

v = Vi - 9.81 t

0 = Vi - 9.81 t at top
Vi = 9.81 t = 19.62 meters/second

Maybe get H as well

falls for five seconds now
h = Hi + Vi T - (1/2) g T^2
0 = Hi + 5 Vi - 4.9(25)
Hi = 122.5 - 96 = 26.5 meters

To determine the initial velocity of the rock, we can use the equations of motion to analyze its vertical motion.

Let's assign some variables:
- Initial velocity (Vi) = ?
- Final velocity (Vf) = 0 m/s (at the highest point and when it hits the ground)
- Acceleration (a) = -9.8 m/s² (downward acceleration due to gravity)
- Time taken to reach the highest point (t₁) = 2 seconds
- Total time of flight (t) = 5 seconds (from throwing the rock until it hits the ground)

First, let's find the velocity at the maximum height (Vmax) using the equation:

Vf = Vi + a * t
0 = Vi - 9.8 * 2

Simplifying the equation, we get:
Vi = 9.8 * 2

Vi = 19.6 m/s

Now, let's calculate the vertical distance the rock travels to reach its maximum height by using the formula:

Δy = Vi * t + (1/2) * a * t²
Δy = 19.6 * 2 + (1/2) * (-9.8) * 2²

Simplifying further:
Δy = 39.2 - 19.6
Δy = 19.6 meters

Since the rock goes up to the same height and then comes back down, the total distance traveled vertically is twice the distance calculated above.

Total vertical distance (H) = 2 * 19.6
H = 39.2 meters

Now, we can use the total time of flight (t) and the total vertical distance (H) to find the initial velocity (Vi) using the equation:

H = (Vi * t) + (1/2) * a * t²
39.2 = Vi * 5 + (1/2) * (-9.8) * 5²

Simplifying the equation, we get:
39.2 = 5Vi - 122.5

Rearranging the equation, we have:
5Vi = 39.2 + 122.5
5Vi = 161.7

Finally, we can solve for Vi:
Vi = 161.7 / 5

Vi ≈ 32.34 m/s

Therefore, the initial velocity of the rock thrown upwards from the edge of the cliff is approximately 32.34 m/s.