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 36P

(a)

To determine

To Find:The range of a ball thrown horizontally while standing on level ground.

(a)

Expert Solution
Check Mark

Answer to Problem 36P

  17.13 m

Explanation of Solution

Given data:

The ball is thrown horizontally.

Formula used:

Second equation of motion:

  s=ut+12at2s

Calculation:

Applying s=ut+12at2 in the horizontal direction,

  R=vt (there is no acceleration in the horizontal direction)

Applying s=ut+12at2 in the vertical downward direction

  h=12gt2t= 2hg

Substituting value of t in R=vt

  R=v×2hg

Assume that the person can throw the ball at 60 mph.

  v=60mph = 60×0.44704ms-1=26.82m s-1

Assume that the person releases the ball from a height of 2 m.

  h=2 m

Substituting the values for v and h

  R=26.82× 2×2 9.81 mR=17.13 m

Conclusion:

Thus, the range of a ball thrown horizontally while standing on level ground = 17.13 m .

(b)

To determine

To Find:The range of a ball thrown at an angle 45° above the horizontal while standing on level ground

(b)

Expert Solution
Check Mark

Answer to Problem 36P

  74.10 m

Explanation of Solution

Given data:

The ball is thrown at an angle 45° above the horizontal.

Formula used:

  s=ut+12at2s

Calculation:

Assume that the person can throw the ball at 60 mph.

  v=60mph = 60×0.44704ms-1=26.82m s-1

Assume that the person releases the ball from a height of 2 m.

  h=2 m

Let the angle of release of the ball from the horizontalbe θ.

Applying s=ut+12at2 in the horizontal direction,

  R=vcosθt

Applying s=ut+12at2 in the vertical downward direction

  h=vsinθt+12gt2

Substituting value of t in R=vcosθt to h=vsinθt+12gt2

  h=vsinθ×Rvcosθ+12g( R vcosθ)22hv2cos2θ=v2Rsin2θ+gR2

Solving for R

  R=v2sin2θ±v4 sin22θ+4g2hv2 cos2θ2g

There are 2 values for R. Taking the larger value,

  R=v2sin2θ+v4 sin22θ+4g2hv2 cos2θ2g

Substituting values for v, h, θ and g

  R= ( 26.82 )2sin( 2× 45 )+ ( 26.82 ) 4 sin 2 ( 2× 45 )+8×9.81×2× ( 26.82 ) 2 cos 2 45 2×9.81 mR=74.10 m

Conclusion:

The range of a ball thrown at an angle 45° above the horizontal = 74.10 m

(c)

To determine

The range of a ball thrown horizontallyfrom the top of a building 12 m high

(c)

Expert Solution
Check Mark

Answer to Problem 36P

The range of a ball thrown horizontally from the top of a building = 45.31 m

Explanation of Solution

Given data:

The ball is thrown horizontallyfrom the top of a building 12 m high.

Formula used:

  s=ut+12at2s

Calculation:

Assume that the person can throw the ball at 60 mph.

  v=60mph = 60×0.44704m.s-1=26.82m.s-1

Assume that the person releases the ball from a height of 2 m.

  h=2 m

Applying s=ut+12at2 in the horizontal direction,

  R=vt

Applying s=ut+12at2 in the vertical downward direction

  h+12=12gt2

Substituting value of t in R=vt to h+12=12gt2

  h+12=12g( R v)2R= 2v( h+12 )g

Substituting values for v, h,and g

  R= 2× ( 26.82 ) 2 ×( 2+12 ) 9.81 mR=45.31 m

Conclusion:

The range of a ball thrown horizontally from the top of a building = 45.31 m

(d)

To determine

The range of a ball thrown at an angle 45° above the horizontal from the top of a building 12 m high

(d)

Expert Solution
Check Mark

Answer to Problem 36P

The range of a ball thrown at an angle 45° above the horizontal from the top of a building 12 m high = 85.35 m

Explanation of Solution

Given data

The ball is thrown at an angle 45° above the horizontal from the top of a building 12 m high.

Formula used

  s=ut+12at2s

Calculation

Assume that the person can throw the ball at 60 mph.

  v=60mph = 60×0.44704m.s-1=26.82m.s-1

Assume that the person releases the ball from a height of 2 m.

  h=2 m

Let the angle of release of the ball from the horizontalbe θ.

Applying s=ut+12at2 in the horizontal direction,

  R=vcosθt

Applying s=ut+12at2 in the vertical downward direction

  h+12=vsinθt+12gt2

Substituting value of t in R=vcosθt to h=vsinθt+12gt2

  h+12=vsinθ×Rvcosθ+12g( R vcosθ)22(h+12)v2cos2θ=v2Rsin2θ+gR2

Solving for R

  R=v2sin2θ±v4 sin22θ+4g×2( h+12)v2 cos2θ2g

There are 2 values for R. Taking the larger value,

  R=v2sin2θ+v4 sin22θ+4g×2( h+12)v2 cos2θ2g

Substituting values for v, h, θ and g

  R= ( 26.82 )2sin( 2× 45 )+ ( 26.82 ) 4 sin 2 ( 2× 45 )+8×9.81×( 2+12 )× ( 26.82 ) 2 cos 2 45 2×9.81 mR=85.35 m

Conclusion

The range of a ball thrown at an angle 45° above the horizontal from the top of a building 12 m high = 85.35 m

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Chapter 3 Solutions

Physics for Scientists and Engineers

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