hapter 03, Problem 37 Your answer is partially correct. Try again. n airplane wi

Question

hapter 03, Problem 37 Your answer is partially correct. Try again. n airplane with a speed of 91.5 m/s ls cir bing upward at an angle of 41.0with respect to the horizontal. When the plane's altitude is 786 m, the pilot releases a ackage. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative the ground, determine the angle of the velocity vector of the package Just before impact. a) Number 131 b) Num 2.9

Explanation / Answer

speed, u=91.5 m/sec

theta=41 degrees,

height, h=786 m

a)

horizontal component of velocity,

ux=u*cos(41)

ux=91.5*cos(41)

ux=69.05 m/sec

and

vertical component of velocity,

uy=u*sin(41)

uy=91.5*sin(41)

uy=60.03 m/sec

and

h=u*t+1/2*a*t^2

h=uy*t+1/2*ay*t^2

-786=60.03*t+1/2*(-9.8)*t^2

===> t=20.2 sec

horizontal distance travelled,

x=ux*t

x=69.05*20.2

x=1394.81 m

b)

final vertical velocity,

vy=uy+ay*t

vy=60.03-9.8*20.2

vy=-137.93 m/sec

and

vx=ux=69.05 m/sec

now,

tan(alpa)=vy/vx

=-137.93/69.05

===> alpa=-63.4 degrees,

therefore,

the angle of velocity vector relative to horizontal is, alpa=63.4 degrees

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