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`
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
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