A helicopter is flying horizontally at 8.1 m/s and an altitude of 17 m when a package of emergency medical supplies is ejected horizontally backward with a speed of 10 m/s relative to the helicopter. Ignoring air resistance, what is the horizontal distance between the package and the helicopter when the package hits the ground?
2007-06-13 09:49:19 · 3 answers · asked by terra_flare_aqua_ciel 1 in Science & Mathematics ➔ Physics
distance fallen = (1/2)a * t^2
17 = (1/2) 9.6 * t^2
t = squareroot(17/4.8)
t = 3.54 seconds
At 10 m/s the horizontal distance behind the helicopter would be 35.4 meters.
2007-06-13 09:55:36 · answer #1 · answered by J C 5 · 0⤊ 0⤋
You need to find the time it takes the package to fall use t=sqrt(2h/g) where h is the height and g is just gravity. This is the time that the helicopter and the package have had to move apart. Then you need to find the speed that they move apart from each other. The package is moving backward at 10m/s relative to the helicopter. So know just use the rate equation
d=v*sqrt(2h/g) you should get 19M
2007-06-13 10:22:10 · answer #2 · answered by ctmtz 2 · 0⤊ 0⤋
2.51 feet
2007-06-13 09:57:45 · answer #3 · answered by Anonymous · 0⤊ 0⤋