A helicopter traveling upward at 106 m/s drops a package from a height of 500 meters. To the nearest second how long does it take to hit the ground?
I'm not sure what numbers to use. What formula to use or anything.. I think 106 m/s is the Vo^2 and 500 m is the h?
I had a similar question that was 120 m/s from 500 meters and got 15.8 seconds but the answer was 28.. Anyone have an idea of how to answer this?
d = Vo*t + 05g*t^2 = 500 m.
-106*t + 4.9t^2 = 500
4.9t^2 -106t - 500 = 0
Use Quad. Formula and get
t = 26 s.
NOTE: The initial velocity is negative , because the helicopter is moving in opposite direction of package.
To find the time it takes for the package to hit the ground, we can use the equation of motion for free fall:
\( h = V_0t + \frac{1}{2}gt^2 \)
where:
h = height (500 m)
V₀ = initial velocity (106 m/s)
t = time
g = acceleration due to gravity (9.8 m/s²)
First, let's rearrange the equation to solve for time (t):
\( h = V_0t + \frac{1}{2}gt^2 \)
Rearranging, we have:
\( \frac{1}{2}gt^2 + V_0t - h = 0 \)
This is a quadratic equation in terms of time (t), so we can solve it using the quadratic formula:
\( t = \frac{-V_0 \pm \sqrt{V_0^2 - 4(\frac{1}{2}g)(-h)}}{2(\frac{1}{2}g)} \)
Substituting the given values:
\( t = \frac{-106 \pm \sqrt{106^2 - 4(\frac{1}{2})(9.8)(-500)}}{2(\frac{1}{2})(9.8)} \)
Evaluating this expression will give us the two possible solutions for time. However, as the question asks for the time to the nearest second, we will consider only the positive solution.
The result will be dependent on the values plugged into the formula. Please evaluate the expression to get the final answer.
To solve this problem, we can use the kinematic equation for vertical motion:
h = Vo*t - (1/2)*g*t^2
Where:
- h is the height (500 meters in this case)
- Vo is the initial vertical velocity (106 m/s in this case)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken to hit the ground (which we need to find)
We can rearrange the equation to solve for t:
h = Vo*t - (1/2)*g*t^2
500 = 106t - (1/2)*9.8*t^2
0 = 4.9t^2 - 106t + 500
Now we can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
In our case, a = 4.9, b = -106, and c = 500. Plugging these values into the formula, we get:
t = (-(-106) ± √((-106)^2 - 4*4.9*500))/(2*4.9)
t = (106 ± √(11236 - 9800))/(9.8)
t = (106 ± √1436)/9.8
Now, calculating the value inside the square root:
√1436 ≈ 37.9
Plugging this back into the equation, we have:
t = (106 ± 37.9)/9.8
Now we have two solutions because of the ± sign:
t1 = (106 + 37.9)/9.8 ≈ 14.6
t2 = (106 - 37.9)/9.8 ≈ 7.0
Since time cannot be negative in this context, the appropriate solution is t ≈ 7.0 seconds. Therefore, to the nearest second, it takes approximately 7 seconds for the package to hit the ground.