Physics, 26.09.2019 11:10 fernandar44

An attacker at the base of a castle wall 3.62 m high throws a rock straight up with speed 8.60 m/s at a height of 1.59 m above the ground. (a) will the rock reach the top of the wall? yes no (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? m/s (c) find the change in the speed of a rock thrown straight down from the top of the wall at an initial speed of 8.60 m/s and moving between the same two points. 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 upwards between the same elevations? explain physically why or why not.

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Answer from: DragonLovely

(a) Yes

We can calculate the total vertical distance covered by the rock by using the equation

v^2 - u^2 = 2ad

where

v = 0 is the final velocity of the rock (the maximum height is reached when v=0)

u = 8.60 m/s is the initial velocity

a = g = -9.8 m/s^2 is the acceleration due to gravity (negative since it points opposite to the velocity)

d is the vertical distance covered

Solving for d,

$d=\frac{v^2 -u^2}{2g}=\frac{0-(8.60 m/s)^2}{2(-9.8 m/s^2)}=3.77 m$

The rock starts its motion from a heigth of

y_0 = 1.59 m

So the maximum height it will reach is

h=y_0 + d=1.59 m +3.77 m =5.36 m

and since the wall is 3.62 m high, this means that the rock will reach the top of the wall.

(b) 5.85 m/s

We can find the velocity of the ball when it reaches the top of the wall by using again the formula

v^2 - u^2 = 2gd

where this time we have

v = ? is the velocity of the rock at the top of the wall

u = 8.60 m/s is the initial velocity

g = -9.8 m/s^2

d = 3.62 m - 1.59 m = 2.03 m is the vertical distance covered by the rock to reach the top of the wall

Solving for v,

$v=\sqrt{u^2 + 2gd}=\sqrt{(8.60 m/s)^2+2(-9.8 m/s^2)(2.03 m)}=5.85 m/s$

Again, we can use the same formula:

v^2 - u^2 = 2gd

But this time we have:

v = ? is the final velocity of the rock at the bottom of the wall

u = 8.60 m/s is the initial velocity of the rock

g = +9.8 m/s^2 , this time positive since the acceleration (downward) has same direction as the velocity (downward)

d = 2.03 m is the distance between the top of the wall and the final point of the motion (1.59 m above the ground)

Solving for v,

$v=\sqrt{u^2 + 2gd}=\sqrt{(8.60 m/s)^2+2(+9.8 m/s^2)(2.03 m)}=10.67 m/s$

So, the change in speed is

$\Delta v = v -u =10.67 m/s - 8.60 m/s = 2.07 m/s$

(d) No

In fact, the change in speed of the rock moving upward is (using the result found in part b):

$\Delta v = v-u = 5.85 m/s - 8.60 m/s = -2.75 m/s$

which we see is different from the change in speed found at part c (rock moving downwards).

This is due to the fact that the rock has an initial velocity different from zero: for this reason, the two situations are not symmetrical:

- when the rock is thrown upward, the gravity acts against the direction of motion, so it makes the rock slowing down

- when the rock is thrown downward, the gravity acts in the same direction as the motion, so it makes the rock speeding up

Answer from: Quest

Hey can someone me with my homework

Answer from: Quest

We have no idea what the graph shows, or why it may be accurate, because we can't see the graph.on a different subject, choice ' d ' is the only true statement among the choices.

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An attacker at the base of a castle wall 3.62 m high throws a rock straight up with speed 8.60 m/s a...

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