COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 4, Problem 71P

(a)

To determine

The time taken by the stone to fall to the base of the gorge.

(a)

Expert Solution
Check Mark

Answer to Problem 71P

The time taken by the stone to fall to the base of the gorge is 3.49s_.

Explanation of Solution

Write the kinematic equation.

    y=vit+12at2        (I)

Here, y is the distance travelled by the stone in the y direction, vi is the initial velocity of the stone, t is the time taken by the stone to fall to the base of the gorge, and a is the acceleration of the stone.

Since the initial velocity of the stone is zero, and the acceleration of the stone is equal to the acceleration due to gravity. Thus the above equation is reduced to

    y=0+12(g)t2t=2yg        (II)

Conclusion:

Substitute 60m for y, and 9.8m/s2 in equation(II), to find t.

    t=2(60m)(9.8m/s2)=3.49s

Therefore, the time taken by the stone to fall to the base of the gorge is 3.49s_.

(b)

To determine

The time taken by the stone to fall to the base of the gorge, if the stone has an initial velocity 20.0m/s2.

(b)

Expert Solution
Check Mark

Answer to Problem 71P

The time taken by the stone to fall to the base of the gorge, if the stone has an initial velocity 20.0m/s2 is 2s_.

Explanation of Solution

Write the kinematic equation.

    y=vit+12at2        (III)

Here, y is the distance travelled by the stone in the y direction, vi is the initial velocity of the stone, t is the time taken by the stone to fall to the base of the gorge, and a is the acceleration of the stone.

Since the initial velocity of the stone is 20.0m/s, the stone travels a distance of 60m in the y direction and the acceleration of the stone is equal to the acceleration due to gravity. Thus the above equation is reduced to

    (60m)=(20m/s)t+12(9.8m/s2)t2        (IV)

Rearrange the above equation.

    (4.9m/s2)t2+(20m/s)t+(60m)=0

Solve the above quadratic equation.

    t=(20m/s)±(20m/s)24(4.9m/s2)(60m)2(4.9m/s2)=(20m/s)±39.7m/s9.8m/s2        (V)

Considering the positive value of time, thus the equation is reduced to

    t=(20m/s)+39.7m/s9.8m/s2=2s

Conclusion:

Therefore, the time taken by the stone to fall to the base of the gorge, if the stone has an initial velocity 20.0m/s2 is 2s_.

(c)

To determine

At what distance the stone will hit the ground from the bridge.

(c)

Expert Solution
Check Mark

Answer to Problem 71P

The stone will hit the ground at a distance of 45.4m_ from the bridge.

Explanation of Solution

Given that the stone is projected at an angle of 30° from the bridge, and the velocity of projection is 20m/s.

Write the expression for horizontal component of velocity.

    vx=(20m/s)cos30°=17.32m/s        (VI)

Here, vx is the horizontal component of velocity.

Write the expression for vertical component of velocity.

    vy=(20m/s)sin30°=10m/s        (VII)

Here, vy is the vertical component of velocity.

Write the kinematic equation.

    y=vyt+12at2        (VIII)

Here, y is the distance travelled by the stone in the y direction, vy is the initial velocity of projection, t is the time taken by the stone to fall to the base of the gorge, and a is the acceleration of the stone.

Since the initial velocity of projection is 10.0m/s, the stone travels a distance of 60m in the y direction and the acceleration of the stone is equal to the acceleration due to gravity. Thus the above equation is reduced to

    (60m)=(10m/s)t+12(9.8m/s2)t2        (IX)

Rearrange the above equation.

    (4.9m/s2)t2+(20m/s)t+(60m)=0

Solve the above quadratic equation.

    t=(10m/s)±(10m/s)24(4.9m/s2)(60m)2(4.9m/s2)=(10m/s)±35.7m/s9.8m/s2        (X)

Considering the positive value of time, thus the equation is reduced to

    t=(10m/s)+35.7m/s9.8m/s2=2.62s

Write the expression for horizontal distance travelled by the stone.

    x=vxt        (XI)

Here, x is the horizontal distance travelled by the stone.

Conclusion:

Substitute 17.32m/s for vx, and 2.62s for t in equation (XI), to find x.

    x=(17.32m/s)(2.62s)=45.4m

Therefore, the stone will hit the ground at a distance of 45.4m_ from the bridge.

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Chapter 4 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

Ch. 4.4 - Prob. 4.8PPCh. 4.5 - Prob. 4.5ACPCh. 4.5 - Prob. 4.10PPCh. 4.5 - Prob. 4.5BCPCh. 4.6 - Prob. 4.11PPCh. 4.6 - Prob. 4.6CPCh. 4 - Prob. 1CQCh. 4 - Prob. 2CQCh. 4 - Prob. 3CQCh. 4 - Prob. 4CQCh. 4 - Prob. 5CQCh. 4 - Prob. 6CQCh. 4 - Prob. 7CQCh. 4 - Prob. 8CQCh. 4 - Prob. 9CQCh. 4 - Prob. 10CQCh. 4 - Prob. 11CQCh. 4 - Prob. 12CQCh. 4 - Prob. 13CQCh. 4 - Prob. 14CQCh. 4 - Prob. 15CQCh. 4 - Prob. 16CQCh. 4 - Prob. 17CQCh. 4 - Prob. 18CQCh. 4 - Prob. 19CQCh. 4 - Prob. 1MCQCh. 4 - Prob. 2MCQCh. 4 - Prob. 3MCQCh. 4 - Prob. 4MCQCh. 4 - Prob. 5MCQCh. 4 - Prob. 6MCQCh. 4 - Prob. 7MCQCh. 4 - Prob. 8MCQCh. 4 - Prob. 9MCQCh. 4 - Prob. 10MCQCh. 4 - Prob. 11MCQCh. 4 - Prob. 12MCQCh. 4 - Prob. 13MCQCh. 4 - Prob. 14MCQCh. 4 - Prob. 15MCQCh. 4 - Prob. 16MCQCh. 4 - Prob. 17MCQCh. 4 - Prob. 18MCQCh. 4 - Prob. 19MCQCh. 4 - Prob. 20MCQCh. 4 - Prob. 21MCQCh. 4 - Prob. 1PCh. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - Prob. 6PCh. 4 - Prob. 7PCh. 4 - Prob. 8PCh. 4 - Prob. 9PCh. 4 - Prob. 10PCh. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19PCh. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 61PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69PCh. 4 - Prob. 70PCh. 4 - Prob. 71PCh. 4 - Prob. 72PCh. 4 - Prob. 73PCh. 4 - Prob. 74PCh. 4 - Prob. 75PCh. 4 - Prob. 76PCh. 4 - Prob. 77PCh. 4 - Prob. 78PCh. 4 - Prob. 79PCh. 4 - Prob. 80PCh. 4 - Prob. 81PCh. 4 - Prob. 82PCh. 4 - Prob. 83PCh. 4 - Prob. 84PCh. 4 - Prob. 85PCh. 4 - Prob. 86PCh. 4 - Prob. 87PCh. 4 - Prob. 88PCh. 4 - Prob. 90PCh. 4 - Prob. 91PCh. 4 - Prob. 92PCh. 4 - Prob. 93PCh. 4 - Prob. 94PCh. 4 - Prob. 95PCh. 4 - Prob. 97PCh. 4 - Prob. 98PCh. 4 - Prob. 99PCh. 4 - Prob. 100PCh. 4 - Prob. 101PCh. 4 - Prob. 102PCh. 4 - Prob. 103PCh. 4 - Prob. 104PCh. 4 - Prob. 105P
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