a photographer inside a helicopter which is descending vertically at 15m/s at an altitude of 55 meters accidentally dropped his camera. how long will it take the camera to reach the ground.
To find the time it takes for the camera to reach the ground, we can use the equation of motion for vertical motion:
s = ut + (1/2)at^2
Where:
- s is the displacement (distance) traveled
- u is the initial velocity
- a is the acceleration
- t is the time
In this case, the camera is dropped, so the initial velocity (u) is 0 m/s, and the acceleration (a) is the acceleration due to gravity, which is approximately 9.8 m/s^2 (assuming no air resistance).
The displacement (s) is the altitude of the helicopter, which is 55 meters.
Therefore, the equation becomes:
55 = (1/2) * 9.8 * t^2
To find the time (t), we can rearrange the equation:
t^2 = (2 * 55) / 9.8
t^2 = 11.22
Taking the square root of both sides:
t ≈ 3.35 seconds
So, it will take approximately 3.35 seconds for the camera to reach the ground.
the free-fall equation is ... h = -1/2 g t^2 - 15 t + 55
... g = 9.8
solve for t ... the positive solution