A parachutist with a camera, both descend- ing at a speed of 11.7 m/s, releases the camera at an altitude of 27.3 m.

What is the magitude of the velocity of the camera just before it hits the ground? The acceleration of gravity is 9.8 m/s2 and air friction is negligible.
Answer in units of m/s

To find the magnitude of the velocity of the camera just before it hits the ground, we can use the equations of motion.

Let's break down the question and the given information:

- The initial velocity of the camera is the same as the parachutist, which is 11.7 m/s.
- The camera is released at an altitude of 27.3 m.
- The acceleration due to gravity is 9.8 m/s^2.
- Air friction is negligible.

We can use the equation of motion, which relates displacement (Δy), initial velocity (v₀), acceleration (a), and final velocity (v):

v² = v₀² + 2aΔy

Substituting the given values into the equation:

v² = (11.7 m/s)² + 2 * (9.8 m/s²) * 27.3 m

Now we can calculate:

v² = 136.89 m²/s² + 534.84 m²/s²

v² = 671.73 m²/s²

To find the magnitude of the velocity (v), we take the square root of v²:

v ≈ √671.73 m²/s²

v ≈ 25.94 m/s

Therefore, the magnitude of the velocity of the camera just before it hits the ground is approximately 25.94 m/s.

h= (v²-v₀²)/2g

solve for ‘v’