a helicopter, travelling horizontally at a constant velocity, is 82m above ground. The pilot prepares to release a package intended to land on the ground 96m horizontally ahead. no air resistance. He lets the package drop. what is the velocity of the helicopter?
The package was released 82 m above ground. With no air resistance, and acceleration due to gravity = 9.81, time it takes is
82 = (1/2)gt²
t=sqrt(2*82/9.81)
= 4.089 s.
During this time, the helicopter has travelled 96 m horizontally, therefore
Velocity of helicopter = 96/4.089=23.48m/s
To find the velocity of the helicopter, we need to analyze the horizontal motion of the package.
Given that the package is dropped from a height of 82m and lands 96m horizontally away, we can consider these distances as displacements in the vertical and horizontal directions, respectively.
In the absence of air resistance, the vertical motion of the package can be described using the equation:
y = yo + vo*t + (1/2)*a*t^2
Where:
y = vertical displacement (82m)
yo = initial height (0m)
vo = initial vertical velocity (0 m/s)
t = time taken
a = acceleration due to gravity (-9.8 m/s^2)
Since the package is dropped, the initial velocity (vo) is 0. Furthermore, the initial height (yo) is chosen as 0 for convenience.
Using the equation, we have:
82 = 0 + 0*t + (1/2)*(-9.8)*t^2
Simplifying the equation:
82 = -4.9t^2
Rearranging the equation:
4.9t^2 = -82
Dividing both sides by 4.9:
t^2 = -82 / 4.9
t^2 ≈ -16.73
Since time cannot be negative for this scenario, we can conclude that there is no solution for this equation. This implies that the package cannot be dropped to land at a point 96m horizontally away while the helicopter is traveling at a constant horizontal velocity.
Therefore, we cannot determine the velocity of the helicopter based on the given information.
To determine the velocity of the helicopter, we need to understand the relationship between distance, time, and velocity.
In this scenario, we are given the horizontal distance (96m) and the vertical distance (82m) that the package will travel. We can use these distances to calculate the time it takes for the package to reach the ground.
First, let's find the time it takes for the package to fall vertically to the ground. We can use the following kinematic equation:
s = ut + (1/2)at^2
Where:
- s is the vertical distance (82m)
- u is the initial vertical velocity (0 m/s since the package is falling vertically)
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time it takes to reach the ground (unknown)
Rearranging the equation, we have:
t = sqrt(2s/a)
Substituting the values, we get:
t = sqrt((2 * 82) / 9.8) = sqrt(16.73) ≈ 4.09 seconds
Now that we have the time, we can calculate the horizontal velocity of the helicopter.
Horizontal velocity can be defined as:
v = d/t
Where:
- v is the horizontal velocity
- d is the horizontal distance (96m)
- t is the time (4.09 seconds)
Substituting the values, we get:
v = 96m / 4.09s ≈ 23.46 m/s
Therefore, the velocity of the helicopter is approximately 23.46 m/s.