3. A helicopter, traveling over level ground, begins to climb. The pilot flies t

Question

3. A helicopter, traveling over level ground, begins to climb. The pilot flies the helicopter with a constant
velocity making an angle of 37 above the horizontal. Then, a package is dropped from the helicopter.
When the package lands 10 s later, it has undergone a horizontal displacement of 120 m. While the
package falls, the velocity of the helicopter is constant.
(a) What is the speed of the helicopter?

(b) How high is the helicopter above ground when the package lands?

(c) How high was the helicopter above ground when the package was dropped?

Explanation / Answer

a) let vo is the initial speed.

here x-component of the veloicty of the package remainse same through out the motion.

vox = x/t

= 120/10

= 12 m/s

we know, vox = vo*cos(37)

==> vo = vox/cos(37)

= 12/cos(37)

= 15.02 m/s <<<<<<--------Answer

c)let h is the initial height.

use the kinematic equation,

-h = voy*t - 0.5*g*t^2

= 15.02*sin(37)*10 - 0.5*9.8*10^2

= -399.7 m

so, h = 399.7m <<<<<<--------Answer for part C)

b) Hmax = h + voy*t

= 417.8 + 15.02*sin(37)*10

=508.2 m <<<<<<--------Answer for part b)

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