a helicopter is ascending vertically with a speed of 5.20 m/s. at a height of 100 m above the Earth, a package is dropped from a window. how much time does it take for the package to reach the ground ? (hint: v0 for the package equals the speed of the helicopter). thank you !
h=v(o) •t +gt^2/2,
g•t^2 +2•v(o) •t -2•h•= 0,
9.8•t^2 +2•5.2 •t -2•100= 0,
t = {-10.4 ±sqrt[(10.4)^2+4•9.8•200]}/2•9.8.
t = 4.01 s
To find the time it takes for the package to reach the ground, we can use the equation of motion for vertical motion:
\(s = u t + \frac{1}{2} a t^2\)
Where:
- \(s\) is the displacement (change in height)
- \(u\) is the initial velocity
- \(a\) is the acceleration (which we assume to be the acceleration due to gravity, approximately \(9.8 \, \text{m/s}^2\) downward for this problem)
- \(t\) is the time
In this case:
- The initial velocity (\(u\)) of the package is equal to the speed of the helicopter, which is 5.20 m/s upward.
- The displacement (\(s\)) is the height at which the package is dropped from, which is 100 m downward (since the package is dropped from above the helicopter).
So we have:
\(100 = (5.20) t + \frac{1}{2} (-9.8) t^2\)
Now we can solve this quadratic equation for \(t\). Rearranging the equation, we get:
\(4.9 t^2 + 5.2 t - 100 = 0\)
We can solve this quadratic equation using the quadratic formula:
\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
For this problem, \(a = 4.9\), \(b = 5.2\), and \(c = -100\).
Plugging in these values in the quadratic formula, we get:
\(t = \frac{-5.2 \pm \sqrt{5.2^2 - 4 \cdot 4.9 \cdot -100}}{2 \cdot 4.9}\)
Simplifying this equation will give us two possible values for \(t\). Since time cannot be negative in this context, we take the positive value.
Solving this equation, we find that \(t = 3.10 \, \text{s}\).
Therefore, it takes approximately 3.10 seconds for the package to reach the ground.
h=v(o) •t +gt^2/2,
g•t^2 +2•v(o) •t -2•h•= 0,
9.8•t^2 +2•5.2 •t -2•100= 0,
t = {-10.4 ±sqrt[(10.4)^2+4•9.8•200]}/2•9.8.
t = 4.01 s