Will a Rock Thrown Upward Reach the Top of a Castle Wall?

[BeckyStar678]

Homework Statement

an attacker at the base of a castle wall 3.65 m high throws a rock straight up with the 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 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 inital 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 teh rock moving upward b/t the same elevations? explain physically why it does or does not agree

Homework Equations

xf=xi+ vxit+1/2axt2

The Attempt at a Solution

so for the first question, what i did was plug in the data for the above equation (making acceleration -9.80) and getting the time which is 9 seconds. so i took that as yes it does reach the top after 9 seconds, because i put xf as 3.65. so then i took the 9 seconds and plugged it in the equation and got 2.4 as the speed at the top. and then i don't know what to do for c.

 

[alphysicist]

Hi BeckyStar678,

Homework Statement

an attacker at the base of a castle wall 3.65 m high throws a rock straight up with the 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 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 inital 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 teh rock moving upward b/t the same elevations? explain physically why it does or does not agree

Homework Equations

xf=xi+ vxit+1/2axt2

The Attempt at a Solution

so for the first question, what i did was plug in the data for the above equation (making acceleration -9.80) and getting the time which is 9 seconds.

I don't believe this is correct; can you show how you got this result (what numbers you used and how you used them in the equation).

 

[baba89]

Your attempt at solving the problem is correct so far. For part c, you can use the same equation and plug in the given initial speed of 7.40 m/s, but this time the acceleration will be positive 9.80 m/s^2, since the rock is now moving downwards. This will give you the time it takes for the rock to reach the ground. Then, you can use the formula for change in speed, which is simply the final speed minus the initial speed. In this case, the final speed will be 0 m/s since the rock will come to a stop at the ground. This will give you the change in speed of the downward-moving rock.

For part d, the change in speed of the downward-moving rock should be the same magnitude as the speed change of the rock moving upward between the same points. This is because the acceleration due to gravity is constant and does not depend on the direction of motion. Therefore, the magnitude of the speed change will be the same for both scenarios.

Physically, this can be explained by the law of conservation of energy. At the top of the wall, the rock has potential energy due to its height, and as it falls, this potential energy is converted into kinetic energy, resulting in an increase in speed. When the rock reaches the ground, all of its potential energy has been converted into kinetic energy, resulting in a change in speed equal to the initial speed. This is true for both the upward and downward motion, resulting in the same magnitude of speed change.