Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 3, Problem 83P

(a)

To determine

The time taken for the stone to reach the base of the gorge.

(a)

Expert Solution
Check Mark

Answer to Problem 83P

The time taken for the stone to reach the base of the gorge is 3.49s .

Explanation of Solution

Write the equation for the time taken for the stone to reach the base of the gorge.

    t=2sg        (I)

Here, t is the time taken for the stone to reach the base of the gorge, s is the height of the gorge and g is the acceleration due to gravity.

Conclusion:

Substitute 60.0m for s and 9.83m/s2 for g in equation (I) to find t.

    t=2(60.0m)9.83m/s2=120.0m9.83m/s2=3.49s

Thus, the time taken for the stone to reach the base of the gorge is 3.49s .

(b)

To determine

The time taken for the stone to reach the ground if it is thrown straight down.

(b)

Expert Solution
Check Mark

Answer to Problem 83P

The time taken for the stone to reach the ground if it is thrown straight down is 2.01s.

Explanation of Solution

Write the equation for the vertical distance.

    y=ut+12gt2        (II)

Here, t is the time taken for the stone to reach the ground, y is the vertical distance, u is the initial velocity and g is the acceleration due to gravity.

Rearrange equation (II),

    12gt2+uty=0        (III)

Conclusion:

Substitute 20.0m/s for u, 60.0m for y and 9.83m/s2 for g in equation (I) to obtain a quadratic equation.

    12(9.83m/s2)t2+(20.0m/s)t60.0m=0

The value of t can be found by solving the above quadratic equation.

  t=b±b24ac2a

Substitute 20.0m/s for b, 1/2 for a, and 60.0m for c and solve.

    t=(20.0m/s)±(20.0m/s)24(12)(9.83m/s2)(60.0m)2(12)(9.83m/s2)=2.01s or6.08s

As the time must be positive, the time taken for the stone to reach the ground if it is thrown straight down is 2.01s.

(c)

To determine

The distance below the bridge the stone will hit the ground.

(c)

Expert Solution
Check Mark

Answer to Problem 83P

The distance below the bridge the stone will hit the ground is 80.6m.

Explanation of Solution

Figure 1 shows the components of velocities.

Physics, Chapter 3, Problem 83P

Figure 1

Write the equation for the vertical distance.

    y=(usinθ)t+12gt2        (III)

Here, t is the time taken for the stone to reach the ground, y is the vertical distance, u is the initial velocity, θ is the projection angle and g is the acceleration due to gravity.

Solve equation (III) ,

  12gt2+(usinθ)ty=0        (IV)

Write the equation for the horizontal distance.

    x=(ucosθ)t        (V)

Conclusion:

Substitute 20.0m/s for u, 60.0m for y, 30.0° for θ and 9.83m/s2 for g in equation (I) to obtain a quadratic equation.

    12(9.83m/s2)t2(20.0m/s)sin30.0°t60.0m=0

Find the value of t by solving the quadratic equation.

  t=b±b24ac2a

Substitute (20.0m/s)sin30.0 for b, 12(9.83m/s2) for a, and 60.0m for c and solve.

    t=b±b24ac2a=(20.0m/s)sin30.0°±(20.0m/s)2sin230.0°4(12)(9.83m/s2)(60.0m)2(12)(9.83m/s2)=4.66s or2.62s

As the time must be positive, the time taken is 4.66s.

Substitute 20.0m/s for u, 30.0s for θ and 4.66s for t in equation (V) to find x.

    x=(20.0m/s)cos30.0°(4.66s)=80.6m

Therefore, the distance below the bridge the stone will hit the ground is 80.6m.

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