An attacker at the base of a castle wall 3.95 m high throws a rock straight up with speed 5.00 m/s from a height of 1.50 m above the ground.

(a) Will the rock reach the top of the wall?

(b) If so, what is its speed at the top? If not, what initial speed must it have to reach the top?

(c) Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 5.00 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 it does or does not agree.

a. h = ho+(V^2-Vo^2)/2g=(0-(5^2))/-19.6

=2.78 m. No.

b. V^2 = Vo^2 + 2g*h
Vo^2 = V^2-2g*h = 0 + 19.6*3.95
Vo^2 = 77.42
Vo = 8.8 m/s.

c. V^2 = Vo^2+2g*h = 5^2 + 19.6*3.95 =
102.42
V = 10.12 m/s.

Change = 10.12-5 = 5.12 m/s.

To answer these questions, we can use the concepts of projectile motion and conservation of energy.

(a) Will the rock reach the top of the wall?
To determine if the rock will reach the top of the wall, we need to compare the maximum height the rock reaches with the height of the wall.

The maximum height reached by the rock can be calculated using the equation:
v^2 = u^2 + 2as, where v is the final velocity (0 m/s at the top), u is the initial velocity (5.00 m/s upwards), a is the acceleration (-9.8 m/s^2 due to gravity), and s is the displacement.

Rearranging the equation to solve for s, we get:
s = (v^2 - u^2) / (2a)

Substituting the values, we have:
s = (0^2 - 5.00^2) / (2(-9.8))
s = (-25) / (-19.6)
s ≈ 1.28 m

The height of the castle wall is 3.95 m, which is greater than the maximum height reached by the rock. Therefore, the rock will not reach the top of the wall.

(b) If not, what initial speed must it have to reach the top?
To find the initial speed required for the rock to reach the top of the wall, we need to determine the maximum height the rock can reach.

Using the same equation as before:
s = (v^2 - u^2) / (2a)

Solving for u (initial velocity) this time:
u^2 = v^2 - 2as
u^2 = 0^2 - 2(-9.8)(3.95)
u^2 = 76.93
u ≈ 8.78 m/s

Therefore, the initial speed of the rock must be approximately 8.78 m/s upwards to reach the top of the wall.

(c) Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 5.00 m/s and moving between the same two points.
When the rock is thrown straight down, it will accelerate due to gravity. The change in speed can be found by subtracting the initial speed from the final speed.

Using the equation:
vf = u + at

Since the rock is moving straight downwards, the acceleration will be positive: a = 9.8 m/s^2

vf = 5.00 + 9.8(3.95 / 9.8) (applying the time it takes to fall the distance of 3.95 m)
vf = 5.00 + 3.95
vf ≈ 8.95 m/s (final velocity)

The change in speed is given by:
change in speed = final velocity - initial velocity
change in speed = 8.95 - 5.00
change in speed ≈ 3.95 m/s

(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 it does or does not agree.
No, the change in speed of the downward-moving rock does not agree with the magnitude of the speed change of the rock moving upward between the same elevations.

Physically, this is because when the rock is thrown straight up, it experiences a deceleration due to the gravitational force pulling it down. The upward speed decreases until it reaches its maximum height and then reverses direction. On the other hand, when the rock is thrown straight down, it accelerates due to the gravitational force, gaining speed as it falls towards the ground. Thus, the change in speed of the downward-moving rock is greater than the magnitude of the speed change of the rock moving upward between the same elevations.