A parachutist drops out of a plane at an altitude of 150m and accidentally drops his camera. If the parachute opens the instant he drops out of the plane and he descents at a constant speed of 6 m/s, how much time separates the arrival of the camera on the ground from the arrival of the parachutist himself?

So I thought to solve the time it took for them separately to reach the ground. I got 5.53 s for the camera. However, for the parachutist I did not know how to solve for it because the initial velocity is 0 and so is the acceleration.
Thank you

Since the camera experiences uniformly accelerated motion (UAM), we can use the formula

h = vo*t - (1/2)gt^2
where
vo = initial velocity (m/s)
h = height (m)
t = time (s)
g = acceleration due to gravity = 9.8 m/s^2

Since the camera is in free fall (it was dropped right before the parachute is opened), its initial velocity is zero. The time it takes for camera to reach the ground is:
150 = 0 - (1/2)(-9.8)t^2
150 = 4.9t^2
t = 5.53 s

Since it was said in the problem that the parachutist falls at a constant speed of 6 m/s,
d = v*t
150 = 6*t
t = 25 s

Therefore,
25 - 5.53 = 19.47 s

hope this helps~ `u`

To solve for the time it takes for the parachutist to reach the ground, you can use the kinematic equation:

Δy = v₀t + (1/2)at²

Where:
Δy = change in vertical displacement (in this case, the height dropped from the plane)
v₀ = initial velocity (which is 0 in this case since the parachutist is initially at rest)
t = time taken
a = acceleration (which is also 0 once the parachute is open, and before that it is the acceleration due to gravity)

Since the parachutist descends at a constant speed of 6 m/s, we can assume that the parachute opens instantly and the parachutist experiences no acceleration before or after opening the parachute. Therefore, a = 0.

Now, let's solve for the time taken for the parachutist to reach the ground using the given information:

Δy = 150 m
v₀ = 0 m/s
a = 0 m/s²

Using the kinematic equation:

Δy = v₀t + (1/2)at²
150 = 0*t + (1/2)*0*t²
150 = 0

As you can see, the equation doesn't allow us to solve for t since it results in an identity. This means that the parachutist will never reach the ground!

However, please note that this is an idealized scenario. In reality, the parachutist will experience air resistance and the opening of the parachute will cause a deceleration due to the drag force. These factors will affect the parachutist's descent, and they need to be taken into account for a more accurate calculation. But given the information provided and the assumption of constant speed, it seems the parachutist will hover at a constant height and never reach the ground.