An Alaskan rescue plane traveling 49 m/s drops a package of emergency rations from a height of 177 m to a stranded party of explorers. The acceleration of gravity is 9.8 m/s 2 . Where does the package strike the ground relative to the point directly below where it was released?
I know the answer is 294.499575551m
But i need to know What is the horizontal component of the velocity just before it hits?
and
What is the vertical component of the velocity just before it hits? (Choose upward as the positive vertical direction)
the horizontal component is constant. If we assume the plane is flying horizontally, then it is always 49 m/s
vertically, v = -9.8t after t seconds. So, how long does it take to hit the ground?
yes. Thus the answer of 294.49 m
To find the horizontal and vertical components of the velocity just before the package hits the ground, we can use the kinematic equations.
First, let's find the time it takes for the package to reach the ground using the formula for vertical motion:
s = ut + (1/2)at^2
Here, s represents the vertical distance traveled (177 m), u is the initial vertical velocity (0 m/s because the package was dropped), a is the acceleration due to gravity (-9.8 m/s^2), and we want to find t (time).
Substituting the values:
177 = 0t + (1/2)(-9.8)(t^2)
Rearranging the equation:
-4.9t^2 = -177
Dividing both sides by -4.9:
t^2 = 36
Taking the square root of both sides:
t = √36
t = 6 s (Since we are solving for time, we discard the negative value)
Now that we have the time it takes for the package to fall, we can find the horizontal and vertical components of the velocity.
The horizontal component of velocity remains constant throughout the motion because no horizontal forces act on the package. Hence, the horizontal component of velocity just before it hits the ground is the same as the initial horizontal velocity, which is given as 49 m/s.
For the vertical component, we can use the formula:
v = u + at
Here, v represents the final vertical velocity just before it hits the ground, u is the initial vertical velocity (0 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time (6 s).
Substituting the values:
v = 0 + (-9.8)(6)
v = -58.8 m/s
Since we are considering upward as the positive vertical direction, we take the magnitude of the velocity:
v = 58.8 m/s
Therefore, the horizontal component of the velocity just before it hits is 49 m/s, and the vertical component of the velocity just before it hits is 58.8 m/s.