a package is released from a helicopter flying horizontally at a constant soeed of 40m/s. the oackage takes 3.0s to reacg the ground. the effects of air resistance can be ignored. what are the horizontal and vertical speeds of the package just before it hits the ground?
There is no force to change the horizontal component of velocity. u = 40, like period. If the helicopter continues at the same speed and heading, the package will hit exactly beneath it. When you drop a bomb, turn :)
It falls for 3 seconds
v = g t = 9.81 * 3 meters/second
Well, let's put on our imaginary helicopter pilot hats and do some calculations! Since the helicopter is flying horizontally, we can assume that its vertical speed is constant and equal to zero because it's not going up or down.
Now, the package takes 3.0 seconds to reach the ground, during which time its vertical speed increases until it reaches the same speed as the helicopter just before impact. Since it starts from rest, we can use the formula v = u + at, where u is the initial velocity (zero), a is the acceleration (which is due to gravity, approximately 9.8 m/s²), and t is the time taken (3.0 seconds).
Using this formula, we find that the vertical speed of the package just before it hits the ground is approximately 29.4 m/s downward. Don't worry, it's just a temporary descent before touchdown!
Since the horizontal speed of the helicopter is constant at 40 m/s, the horizontal speed of the package just before it hits the ground is also 40 m/s. It's like a speedy delivery by air!
So, to summarize, the package's horizontal speed remains at 40 m/s, while its vertical speed increases to approximately 29.4 m/s downward before booping the ground.
To determine the horizontal and vertical speeds of the package just before it hits the ground, we can use the equations of motion.
First, let's consider the vertical motion of the package. We can use the equation:
**Vertical distance (y) = Initial vertical velocity (V₀y) × Time (t) + 0.5 × Acceleration due to gravity (g) × Time squared (t²)**
Since the package is dropped from a helicopter, its initial vertical velocity (V₀y) is 0 m/s (because it is not given an initial upward or downward velocity). We can substitute the known values into the equation:
Vertical distance (y) = (0 m/s) × (3.0 s) + 0.5 × (9.8 m/s²) × (3.0 s)²
Simplifying the equation gives us:
Vertical distance (y) = 0 + 0.5 × 9.8 m/s² × 9.0 s²
= 0 + 44.1 m
= 44.1 m
Since the package hits the ground, the vertical distance covered is equal to the height from which it is dropped.
Now, let's consider the horizontal motion of the package. The horizontal speed remains constant at 40 m/s since there are no horizontal forces acting on the package. Therefore, the horizontal speed just before it hits the ground is also 40 m/s.
To summarize:
- The vertical speed just before the package hits the ground is 0 m/s (since it is in freefall).
- The horizontal speed just before the package hits the ground is 40 m/s.
31
horizontal velocity (m/s) vertical velocity (m/s)
A 0 30
B 30 30
C 30 40
D 40 30
E 40 40
CHOOSE.