Chapter 2, Problem 48P

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

Chapter
Section

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

# An attacker at the base of a castle wall 3.65 m high throws a rock straight up with speed 7.40 m/s at a height of 1.55 m above the ground, (a) Will the rock reach the top of the wall? (b) If so, what is the rock's speed at the top? If not, what initial speed must the rock have to reach the top? (c) Find the change in the speed of a rock thrown straight down from the top of the wall at an initial speed of 7.40 m/s and moving between the same two points, (d) Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? Explain physically why or why not.

(a)

To determine
The possibility of the rock to reach at the top.

Explanation

Given Info:

The initial velocity of the rock is 7.40â€‰m/s .

The initial height from which the rock thrown is 1.55â€‰m .

The acceleration of the rock is âˆ’9.80â€‰m/s2 .

The final velocity of the rock is 0 .

The height of the wall is 3.65â€‰m .

Explanation:

The formula used to calculate the maximum height the rock raises is,

vf2=vi2+2aÎ”y

Here,

Î”y is the displacement of the rock

vf is the final velocity of the rock

vi is the initial velocity of the rock

a is the acceleration

Substitute (hmaxâˆ’hi) for Î”y to find vf2 .

vf2=vi2+2a(hmaxâˆ’hi)

Here,

hmax is maximum height the rock reaches

hi is the initial height from which the rock thrown

Rearrange the equation in terms of hmax

(b)

To determine
The speed of the rock at the top of the wall.

(c)

To determine
The change in speed between the rocks.

(d)

To determine
The possibility of the rocks having the same magnitude of speed change.

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