EXAMPLE 3.5Stranded Explorers From the point of view of an observer on the groun

Question

EXAMPLE 3.5Stranded Explorers

From the point of view of an observer on the ground, a package released from the rescue plane travels along the path shown.

SOLUTION

(A) Find the range of the package.

Use the equation to find the y-displacement.

y = y y0 = v0yt

gt2

Substitute y0 = 0 and v0y = 0, set y = 1.00 = 102 m, the final vertical position of the package relative the airplane. Solve for time.

t = 4.52 s

Use the equation to find the x-displacement.

x = x x0 = v0xt

Substitute x0 = 0, v0x = 40.0 m/s, and the time.

x = (40.0 m/s)(4.52 s) = 181 m

(B) Find the components of the package's velocity at impact.

Find the x-component of the velocity at the time of impact.

vx = v0cos = (40.0 m/s) cos 0° = 40.0 m/s

Find the y-component of the velocity at the time of impact.

(C) Find the angle of the impact.

Write the equation and substitute values.

tan =

=

= 1.11

Apply the inverse tangent function to both sides.

= tan1(1.11) = 48.0°

LEARN MORE

REMARKS Notice how motion in the x-direction and motion in the y-direction are handled separately.

QUESTION Neglecting air friction effects, what path does the package travel as observed by the pilot?

a straight line sloped downwarda downward curved path whose tangent line at each point has negative slope    a vertical line downwarda curved path that starts horizontal and then is increasingly bent downwarda curved path that is first curved upward and then downward

PRACTICE IT

(c) Find the angle of the impact.
°

EXERCISEHINTS:  GETTING STARTED  |  I'M STUCK!

A bartender slides a beer mug at 1.9 m/s towards a customer at the end of a frictionless bar that is 1.1 m tall. The customer makes a grab for the mug and misses, and the mug sails off the end of the bar.(a) How far away from the end of the bar does the mug hit the floor?
m

(b) What are the speed and direction of the mug at impact?

Explanation / Answer

A) lets find time taken,

Using second equation of motion,

- 120 = - 4.9t^2

t = sqrt (120/4.9)

= 4.95 s

Horizontal distance = Vx*t = 39*4.95 = 193 m

B) horizontal velocity =Vx = 39 m/s

Vertical velocity = Vy= 0-9.8*4.95 = - 48.5 m/s

C) angle of impact = arctan (-48.5/39) = - 51.2 degree

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