Physics Fundamentals
Physics Fundamentals
2nd Edition
ISBN: 9780971313453
Author: Vincent P. Coletta
Publisher: PHYSICS CURRICULUM+INSTRUCT.INC.
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Chapter 3, Problem 37P

(a)

To determine

To find: The horizontal distance travelled by the shot for θ=0 .

(a)

Expert Solution
Check Mark

Answer to Problem 37P

  x=5.00m

Explanation of Solution

Given data:

  Initialvelocity(u)=8.00ms-1Height=1.90mAngle =0

Formula used:

  h=12g.t2

  x=u.t

Calculation:

  Physics Fundamentals, Chapter 3, Problem 37P , additional homework tip  1h=12g.t21.90m=12×9.81ms-2.t2t2=1.90m×29.81 ms -2t2=0.38s2t=0.622s

  x=u.tx=8.00ms-1×0.622sx=5.00m

Conclusion:The horizontal distance travelled by the shot for θ=0 is,

  x=5.00m

(b)

To determine

To find: The horizontal distance traveled by the shot for θ=40 .

(b)

Expert Solution
Check Mark

Answer to Problem 37P

  X=9.85 m

Explanation of Solution

Given data:

  Initialvelocity(u)=8.00ms-1Height=1.90mAngle =40

Formula used:

  Vx=V.cosθ

  Vy=V.sin40

  VFinal=VInitialg.t

  h=Vyt+12g.t2

  h=12g.t2

  X=Vxt

Calculation:

  Physics Fundamentals, Chapter 3, Problem 37P , additional homework tip  2

  Vx=V.cosθVx=8m/s.cos40Vx=6.12ms-1

  Vy=V.sin40Vy=8m/s.sin40Vy=5.14ms-1

  VFinal=VInitialg.t0=5.14ms-19.81ms-2.tt=0.52s

  h=Vyt+12g.t2h=(5.14×0.52)12.9.81.(0.52s)2h=1.35m

  hmax=1.35m+1.9mhmax=3.25m

Free fall from the maximum height.

  h=12g.t23.25m=12×9.81×t2t2=.6625t=.6625t=0.814s

  tTotal=0.814s+0.52stTotal=1.334s

  X=VxtX=6.12ms-1×1.334sX=8.16 m

Conclusion:

The horizontal distance traveled by the shot for θ=40 .

  X=8.16 m

(c)

To determine

To find: The horizontal distance traveled by the shot for θ=45 degrees.

(c)

Expert Solution
Check Mark

Answer to Problem 37P

  X=9.83 m

Explanation of Solution

Given data:

  Initialvelocity(u)=8.00ms-1Height=1.90mAngle =45

Formula used:

  Vx=V.cosθ

  Vy=V.sin45

  VFinal=VInitialg.t

  h=Vyt+12g.t2

  h=12g.t2

  X=Vxt

Calculation:

  Physics Fundamentals, Chapter 3, Problem 37P , additional homework tip  3

  Vx=V.cosθVx=8m/s.cos45Vx=5.65ms-1

  Vy=V.sin45Vy=8m/s.sin45Vy=5.65ms-1

  VFinal=VInitialg.t0=5.65ms-19.81ms-2.tt=0.57s

  h=Vyt+12g.t2h=(5.65×0.57)12.9.81.(0.57s)2h=1.63m

  hmax=1.63m+1.9mhmax=3.53m

Free fall from the maximum height.

  h=12g.t23.53m=12×9.81×t2t=.848s

  tTotal=.848s+0.57stTotal=1.42s

  X=VxtX=5.65ms-1×1.42sX=8 m

Conclusion:

The horizontal distance traveled by the shot for θ=45 degrees is,

  X=8.0m

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