An Alaskan rescue plane traveling 36 m/s drops a package of emergency rations from a height of 162 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?
Answer in units of m
What is the horizontal component of the velocity just before it hits?
Answer in units of m/s
What is the vertical component of the velocity just before it hits? (Choose upward as the positive vertical direction)
Answer in units of m/s
h=gt²/2
t=sqrt(2h/g)
L=v•t=36•sqrt(2•162/9.8)=…
v(x)= 36 m/s
v(y)=gt=g•sqrt(2h/g)=...
To answer these questions, we can use the equations of motion to analyze the vertical and horizontal components of the package's motion.
1. Where does the package strike the ground relative to the point directly below where it was released?
The vertical position of the package can be calculated using the equation:
y = y0 + v0y*t + (1/2) * g * t^2
Where:
y = vertical position of the package
y0 = initial vertical position (162 m in this case)
v0y = initial vertical velocity (0 m/s since the package is dropped, not thrown)
g = acceleration due to gravity (-9.8 m/s^2, taking downward as the positive direction)
t = time
Solving for t when the package reaches the ground (y = 0), we can rearrange the equation to:
0 = y0 + v0y*t + (1/2) * g * t^2
This equation is a quadratic equation, so we can solve for t using the quadratic formula:
t = (-v0y ± √(v0y^2 - 2 * g * y0)) / g
Substituting the given values into the equation, we have:
v0y = 0 m/s
g = -9.8 m/s^2
y0 = 162 m
t = (-0 ± √(0^2 - 2 * (-9.8) * 162)) / -9.8
Solving this equation, we get two possible values for t: t1 and t2.
2. What is the horizontal component of the velocity just before it hits?
The horizontal motion of the package is unaffected by gravity. The horizontal velocity remains constant throughout the drop. Therefore, the horizontal component of the velocity just before it hits is the same as it was when the package was dropped. The horizontal velocity is given as 36 m/s.
3. What is the vertical component of the velocity just before it hits?
The vertical velocity of the package at any moment can be calculated using the equation:
v = v0y + g * t
Where:
v = vertical velocity
v0y = initial vertical velocity (0 m/s since the package is dropped)
g = acceleration due to gravity (-9.8 m/s^2, taking downward as the positive direction)
t = time
Using the values calculated for t, we can substitute them into the equation to find the vertical component of velocity just before it hits.