A photographer in a helicopter ascending vertically at a constant rate of 1.2m/s accidentally drops a camera out of the window when the helicopter is 46m above the ground. How long will it take the camera to reach the ground
Hi = 46
Vi = 1.2
a = g = -9.81 m/s^2
h = Hi + Vi t - 4.9 t^2
0 = 46 + 1.2 t - 4.9 t^2
t^2 - .245 t - 9.39 = 0
t = [ .245 +/- sqrt ( .06 + 37.6) ]/2
t = [ .245 + 6.13 ] /2 forget negative time
t = 3.2 seconds
To find the time it takes for the camera to reach the ground, we can use the equation of motion:
h = ut + (1/2)gt^2
Where:
h = height (46m)
u = initial velocity (0 m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = time (unknown)
Since the camera drops from rest, the initial velocity u is 0. The acceleration due to gravity, g, is negative because it acts in the opposite direction of the motion. Therefore, we can rewrite the equation as:
h = (1/2)gt^2
Substituting the known values:
46 = (1/2)(-9.8)t^2
Rearranging the equation:
t^2 = (46 / (1/2)(-9.8))
t^2 = (46 / -4.9)
t^2 = -9.3878
Since time cannot be negative, we discard the negative value:
t = ā(9.3878)
t ā 3.06 seconds
Therefore, it will take approximately 3.06 seconds for the camera to reach the ground.
To find the time it takes for the camera to reach the ground, we can use the kinematic equation for vertical motion:
š = š£āš” + 0.5šš”Ā²
Where:
š = displacement (in this case, the initial height of the camera, which is 46m and the final height, which is 0m)
š£ā = initial velocity (the velocity at which the camera is dropped from the helicopter, which is 0m/s)
š” = time (what we are trying to find)
š = acceleration (the acceleration due to gravity, which is approximately 9.8m/sĀ², downward direction)
Rearranging the equation, we get:
0 = 0 + 0.5(9.8)š”Ā²
Simplifying further:
4.9š”Ā² = 46
Dividing both sides by 4.9:
š”Ā² = 9.387755102
Taking the square root of both sides:
š” ā 3.06 seconds (rounded to two decimal places)
Therefore, it will take approximately 3.06 seconds for the camera to reach the ground.