A helicopter is ascending vertically with a speed of 5.00 m/s. At a height of 120 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?
v_0 for the package equals the speed of the helicopter
Help?
A helicopter is ascending vertically at the rate of 19.6 m/s. When it is at the height of 156.8 m above the ground a stone is dropped. how long does the stone take to reach the ground?
huhrensohn
To find the time it takes for the package to reach the ground, we need to consider the vertical motion of the package. The package is dropped from a height of 120 m, so its initial velocity is zero. The acceleration due to gravity, denoted as "g," is approximately 9.8 m/s².
To solve for the time it takes to reach the ground, we can use the equation:
s = v₀t + (1/2)gt²
Where:
- s is the displacement (in this case, the height) of the package,
- v₀ is the initial velocity of the package,
- t is the time, and
- g is the acceleration due to gravity.
Since the package is dropped from rest, v₀ is 0. The equation then simplifies to:
s = (1/2)gt²
Plugging in the given values:
s = 120 m
g = 9.8 m/s²
120 = (1/2)(9.8)t²
Now we can solve for t by rearranging the equation:
t² = (2s) / g
t = √[(2s) / g]
Substituting the values:
t = √[(2 * 120) / 9.8]
Calculating this expression, we find that t ≈ 5.16 seconds.
Therefore, it takes approximately 5.16 seconds for the package to reach the ground.
Package height Y at drop time t is
Y = 120 + 5 t - 4.9 t^2
Set Y = 0 and solve.