Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 3, Problem 122P
To determine

The time at which the packet fell off the back ramp.

Expert Solution & Answer
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Explanation of Solution

Given: The speed of the airplane is 275mi/hr .

The angle is 37° above the horizontal.

The distance is 7.5km .

Formula used: Draw the diagram for the representation of the package when it fell off the ramp.

  Physics for Scientists and Engineers, Chapter 3, Problem 122P

Write the expression for the distance in x direction.

  xd=vx,dtd

Here, xd is the distance in the x direction, vx,d is the velocity in the x direction and td is the time.

Write the expression for the distance in y direction.

  yd=vy,dtd

Here, yd is the distance in the y direction, vy,d is the velocity in the y direction and td is the time.

Write the expression for the x coordinate along the parabolic trajectory of the airplane.

  x(t)=xd+vx,d(ttd)   ...... (1)

Here, x(t) is the position as the function of time, vx,d is the velocity and td is the time.

Write the expression for the y coordinate along the parabolic trajectory of the airplane.

  y(t)=yd+vy,d(ttd)12g(ttd)2   ...... (2)

Here, y(t) is the position as the function of time, g is acceleration due to gravity.

Substitute vx,dtd for xd in equation (1).

  x(t)=vx,dtd+vx,d(ttd)

Substitute vy,dtd for yd in equation (2).

  y(t)=vy,dtd+vy,d(ttd)12g(ttd)2

Simplify the above equation

  y(t)=vy,dtd12g(ttd)2

Write the expression for the velocity component with respect to x axis.

  vx,d=vpcosθ0   ...... (3)

Here, vp is the velocity of the plane and θ0 is the angle.

Write the expression for the velocity component with respect to y axis.

  vy,d=vpsinθ0   ...... (4)

At impact, the horizontal distance is xl and vertical distance is zero.

Write the expression for the distance to impact.

  xl=vx,dtt

Simplify the above equation for tt .

  tt=xlvx,d

Substitute xlvx,d for tt and 0 for y(t) in equation (2).

  0=vy,dtd12g( x l v x,d td)2

Substitute xlvx,d for td in the above equation.

  0=vy,dxlvx,d12g( x l v x,d td)2

Solve the above equation using binomial expansion.

  td22xlvx,dtd+(xl2v x,d22xlv y,dgv x,d)=0   ...... (5)

Calculation:

Substitute 275mi/hr for vp and 37° for θ0 in equation (3).

  vx,d=(275 mi hr×1609m mi× 1hr 3600s)cos37°vx,d=98.16m/s

Substitute 275mi/hr for vp and 37° for θ0 in equation (4).

  vy,d=(275 mi hr×1609m mi× 1hr 3600s)sin37°vy,d=73.97m/s

Substitute 7.5km for xl , 73.97m/s for vy,d , 98.16m/s for vx,d and 9.81m/s2 for g in equation (5).

  td22( 7.5km)98.16m/std+( ( 7.5km ) 2 ( 98.16m/s ) 2 2( 7.5km )73.97m/s ( 98.16m/s ))=0td2(152.8s)td+4686s2td=42s

Conclusion:

Thus, the time at which the packet fell off the back ramp is 42s .

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Chapter 3 Solutions

Physics for Scientists and Engineers

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