A rock is thrown vertically from the edge of a cliff that is 500 ft high. It takes 6.5 seconds for the rock to hit the ground at the base of the cliff. Assume air resistance is negligible.

(a) What was the initial velocity (magnitude and direction) of the rock?
(b) What was the magnitude of the velocity of the rock just before impact with the ground?

(a) They have already told you that the initial rock direction was vertical (up).

y = 500 + Vo*t - 16.1 t^2

= 0 then t = 6.5

0 = 500 + 6.5*Vo - 16.1*(6.5)^2

Solve for Vo.

V(t) = Vo - 16.1 t
Substitute t = 6.5 s and solve for V at that time.

The number 16.1 is the value of g/2, in units of ft/s^2

youve got the wrong answer

To answer these questions, we can use the equations of motion for a vertically thrown object under constant acceleration due to gravity. Here are the steps to find the answers:

(a) What was the initial velocity (magnitude and direction) of the rock?

Step 1: Identify the known variables:
- Initial position (s₀) = 500 ft
- Time (t) = 6.5 s

Step 2: Find the acceleration (a):
The acceleration due to gravity is constant and points downward. We can use the equation s = s₀ + v₀t + (1/2)at², where s is the final position.

Since the rock falls from a height of 500 ft and reaches the ground (s = 0), we can rewrite the equation as 0 = 500 - (1/2)gt². Solving for a, we get a = g = 32.2 ft/s² (approximately).

Step 3: Calculate the initial velocity (v₀):
We can use the equation v = v₀ + at, where v is the final velocity (0 ft/s) at the time of impact.

Rewriting the equation as 0 = v₀ + (32.2 ft/s²)(6.5 s) and solving for v₀, we find v₀ = -208.3 ft/s.

The magnitude of the initial velocity is 208.3 ft/s, and since it's negative, the direction is downward.

(b) What was the magnitude of the velocity of the rock just before impact with the ground?

Since we already know the acceleration (a = -32.2 ft/s²) and the time to impact (t = 6.5 s), we can use the equation v = v₀ + at again to find the velocity just before impact.

Plugging in the values, we get v = -208.3 ft/s + (-32.2 ft/s²)(6.5 s). Solving, we find the magnitude of the velocity just before impact to be 314.3 ft/s.