You want to drop a bundle of all your class notes, tests, quizzes, worksheets, and term papers on the 50 yard-line of Lynn C. Adams field during half-time of the first playoff game from a single engine Piper Cub. You fly at a steady height of 491.0 meters and at a speed of 65.0 m/s. How long will it take for the bundle to reach the ground? And how far in front of the 50 yard-line must the bundle be dropped?
Vertical problem first
Hi = 491 meters, Vi = 0
h = 491 + 0 t - 4.9 t^2
at h = 0
4.9 t^2 = 491
t^2 = 100
t = 10 seconds falling
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horizontal problem,
go x meters in 10 seconds at 65 meters/ second
65 * 10 = 650 meters
To determine the time it takes for the bundle to reach the ground, we can use the equation of motion:
h = (1/2) * g * t^2
Where:
h = height (491.0 meters)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the equation, we get:
t = sqrt(2h / g)
Let's substitute the given values into the equation to find the time:
t = sqrt(2 * 491.0 / 9.8)
= sqrt(982 / 9.8)
= sqrt(100)
= 10 seconds
Therefore, it will take 10 seconds for the bundle to reach the ground.
To determine how far in front of the 50 yard-line the bundle must be dropped, we can use the equation of motion:
d = v * t
Where:
d = distance
v = velocity (65.0 m/s)
t = time (10 seconds)
Let's substitute the given values into the equation to find the distance:
d = 65.0 * 10
= 650 meters
Therefore, the bundle must be dropped 650 meters in front of the 50 yard-line.